Prime Hypotheses, Exact Modular Powers and Failed Fermat Converses
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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Fermat's little theorem · qf-mod37-flt-converse-counterexample · candidate
Canonical question JSONL · English JSONL · Validation report