# Prime Hypotheses, Exact Modular Powers and Failed Fermat Converses

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations · 2026-10-14 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=3 and modulus=2, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(2, 1, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (2, 1, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0001/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 2. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-4 and modulus=3, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(3, 2, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (3, 2, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0002/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 3. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=4 and modulus=5, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(5, 4, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (5, 4, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0003/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 4. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-5 and modulus=7, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(7, 2, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (7, 2, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0004/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 5. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=6 and modulus=11, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(11, 6, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (11, 6, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0005/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 6. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-7 and modulus=13, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(13, 6, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (13, 6, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0006/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 7. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=8 and modulus=17, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(17, 8, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (17, 8, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0007/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 8. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-9 and modulus=19, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(19, 10, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (19, 10, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0008/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 9. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=10 and modulus=23, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(23, 10, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (23, 10, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0009/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 10. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-11 and modulus=29, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(29, 18, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (29, 18, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0010/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 11. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=12 and modulus=31, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(31, 12, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (31, 12, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0011/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 12. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-13 and modulus=37, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(37, 24, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (37, 24, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0012/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 13. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=14 and modulus=41, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(41, 14, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (41, 14, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0013/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 14. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-15 and modulus=43, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(43, 28, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (43, 28, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0014/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 15. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=16 and modulus=47, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(47, 16, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (47, 16, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0015/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 16. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-17 and modulus=53, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(53, 36, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (53, 36, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0016/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 17. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=18 and modulus=59, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(59, 18, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (59, 18, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0017/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 18. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-19 and modulus=61, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(61, 42, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (61, 42, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0018/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 19. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=20 and modulus=67, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(67, 20, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (67, 20, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0019/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 20. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-21 and modulus=71, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(71, 50, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (71, 50, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0020/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 21. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=22 and modulus=73, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(73, 22, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (73, 22, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0021/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 22. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-23 and modulus=79, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(79, 56, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (79, 56, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0022/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 23. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=24 and modulus=83, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(83, 24, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (83, 24, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0023/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 24. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=-25 and modulus=89, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(89, 64, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (89, 64, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0024/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 25. Use the Prime-Coprime Fermat Form Only under Both Hypotheses

Use integer base a=26 and modulus=97, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).

View answer and explanation(97, 26, 1, Yes, prime_coprime_form)

Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (97, 26, 1, Yes, prime_coprime_form).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0025/)

Fermat's little theorem · qf-mod37-flt-prime-coprime · candidate





## 26. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=2 and modulus=2, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(2, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (2, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0026/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 27. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-6 and modulus=3, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(3, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0027/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 28. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=15 and modulus=5, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(5, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (5, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0028/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 29. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-28 and modulus=7, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(7, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (7, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0029/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 30. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=55 and modulus=11, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(11, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (11, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0030/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 31. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-78 and modulus=13, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(13, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (13, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0031/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 32. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=119 and modulus=17, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(17, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (17, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0032/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 33. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-152 and modulus=19, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(19, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (19, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0033/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 34. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=207 and modulus=23, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(23, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (23, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0034/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 35. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-290 and modulus=29, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(29, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (29, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0035/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 36. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=341 and modulus=31, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(31, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (31, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0036/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 37. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-444 and modulus=37, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(37, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (37, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0037/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 38. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=533 and modulus=41, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(41, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (41, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0038/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 39. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-602 and modulus=43, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(43, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (43, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0039/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 40. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=705 and modulus=47, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(47, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (47, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0040/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 41. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-848 and modulus=53, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(53, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (53, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0041/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 42. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=1003 and modulus=59, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(59, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (59, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0042/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 43. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-1098 and modulus=61, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(61, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (61, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0043/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 44. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=1273 and modulus=67, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(67, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (67, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0044/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 45. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-1420 and modulus=71, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(71, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (71, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0045/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 46. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=1533 and modulus=73, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(73, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (73, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0046/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 47. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-1738 and modulus=79, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(79, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (79, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0047/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 48. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=1909 and modulus=83, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(83, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (83, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0048/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 49. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=-2136 and modulus=89, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(89, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (89, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0049/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 50. The All-Integer Prime Form Includes Bases Divisible by p

Use integer base a=2425 and modulus=97, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).

View answer and explanation(97, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)

Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (97, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0050/)

Fermat's little theorem · qf-mod37-flt-all-integer · candidate





## 51. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=2 and modulus=4, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(4, 2, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (4, 2, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0051/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 52. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-3 and modulus=5, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(5, -3, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (5, -3, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0052/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 53. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=4 and modulus=6, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(6, 4, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (6, 4, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0053/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 54. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-5 and modulus=7, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(7, -5, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (7, -5, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0054/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 55. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=6 and modulus=8, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(8, 6, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (8, 6, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0055/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 56. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-7 and modulus=9, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(9, -7, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (9, -7, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0056/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 57. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=8 and modulus=10, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(10, 8, No, No, No, 8, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (10, 8, No, No, No, 8, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0057/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 58. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-9 and modulus=11, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(11, -9, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (11, -9, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0058/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 59. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=10 and modulus=12, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(12, 10, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (12, 10, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0059/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 60. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-11 and modulus=13, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(13, -11, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (13, -11, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0060/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 61. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=12 and modulus=14, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(14, 12, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (14, 12, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0061/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 62. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-13 and modulus=15, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(15, -13, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (15, -13, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0062/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 63. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=14 and modulus=16, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(16, 14, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (16, 14, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0063/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 64. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-15 and modulus=17, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(17, -15, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (17, -15, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0064/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 65. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=16 and modulus=18, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(18, 16, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (18, 16, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0065/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 66. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-17 and modulus=19, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(19, -17, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (19, -17, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0066/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 67. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=18 and modulus=20, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(20, 18, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (20, 18, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0067/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 68. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-19 and modulus=21, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(21, -19, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (21, -19, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0068/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 69. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=20 and modulus=22, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(22, 20, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (22, 20, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0069/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 70. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-21 and modulus=23, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(23, -21, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (23, -21, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0070/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 71. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=22 and modulus=24, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(24, 22, No, No, No, 16, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (24, 22, No, No, No, 16, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0071/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 72. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-23 and modulus=25, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(25, -23, No, Yes, No, 16, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (25, -23, No, Yes, No, 16, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0072/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 73. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=24 and modulus=26, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(26, 24, No, No, No, 24, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (26, 24, No, No, No, 24, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0073/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 74. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=-25 and modulus=27, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(27, -25, No, Yes, No, 13, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (27, -25, No, Yes, No, 13, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0074/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 75. An Inapplicable Theorem Need Not Mean a Failed Modular Test

Use integer base a=26 and modulus=28, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).

Failure of the prime-coprime hypotheses does not force this residue to differ from1; a composite can pass a Fermat test for a particular base.

View answer and explanation(28, 26, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses)

Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (28, 26, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0075/)

Fermat's little theorem · qf-mod37-flt-hypothesis-audit · candidate





## 76. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=2 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 2, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 2, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0076/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 77. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=4 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 4, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 4, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0077/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 78. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=5 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 5, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 5, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0078/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 79. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=7 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 7, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 7, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0079/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 80. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=8 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 8, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 8, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0080/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 81. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=10 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 10, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 10, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0081/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 82. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=13 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 13, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 13, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0082/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 83. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=14 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 14, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 14, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0083/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 84. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=16 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 16, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 16, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0084/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 85. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=19 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 19, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 19, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0085/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 86. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=20 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 20, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 20, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0086/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 87. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=23 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 23, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 23, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0087/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 88. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=25 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 25, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 25, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0088/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 89. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=26 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 26, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 26, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0089/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 90. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=28 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 28, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 28, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0090/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 91. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=29 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 29, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 29, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0091/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 92. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=31 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 31, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 31, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0092/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 93. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=32 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 32, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 32, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0093/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 94. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=35 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 35, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 35, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0094/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 95. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=37 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 37, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 37, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0095/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 96. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=38 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 38, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 38, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0096/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 97. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=40 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 40, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 40, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0097/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 98. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=41 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 41, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 41, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0098/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 99. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=43 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 43, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 43, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0099/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate





## 100. A Composite Can Pass a Fermat Test for a Chosen Base

Use integer base a=46 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).

View answer and explanation(561, 46, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)

Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 46, No, Yes, 1, converse_false_composite_can_pass_Fermat_test).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261106-r05-fermat-little-theorem-0100/)

Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations/validation-report.json)