Two independent routes for Euler's criterion

Goal and scope

For p=7,a=-4 the Legendre value is -1 and a^3=6=-1 mod 7; 2 gives +1.

A complete square table and independent repeated-squaring route, certifying p=7 instances only.

Givens and symbols

• calc

• G1 — p=7,a=-4; normalize to 3 modulo 7, then independently compute the Legendre value and a^((p-1)/2).

• Q_7 — complete nonzero squares
• (a/7) — signed Legendre value

Rules and sources

Some rule statements have reader clarifications in this revision: they are paraphrases, not verbatim quotations. Referenced original source bytes and hashes are unchanged.

• calc · prerequisite — Assume exact integer arithmetic, division with remainder, and definitions of divisibility and gcd; every congruence has an explicit modulus. Assume the square-root trial-division primality criterion; define Legendre values by zero and the complete nonzero square table.
• mko_definition · cited — Require odd primality and a normalized nonzero class first. Compute the Legendre value independently from the complete nonzero square table, and the modular half power by repeated squaring; interpret p−1 as signed−1. Residues give +1 and nonresidues−1. A finite table checks its specific p,a, not the theorem for every odd prime. Modulus2, composite moduli and the zero class are outside this node domain. Do not circularly define Legendre by this criterion or infer a general composite-modulus converse. · mko-euler-criterion-quadratic-residue

Main derivation

p=7,\quad a=-4

7\text{ odd prime},\quad a\equiv3\ne0\pmod7,\quad(p-1)/2=3

x=1,2,3,4,5,6

\begin{array}{c|rrrrrr}x&1&2&3&4&5&6\\\hline x^2\bmod7&1&4&2&2&4&1\end{array}

\begin{array}{c|rrrrrr}x&1&2&3&4&5&6\\\hline x^2\bmod7&1&4&2&2&4&1\end{array}

Q_7=\{1,2,4\},\quad(3/7)=-1,\quad(2/7)=1

a\equiv3\pmod7

3^2=9\equiv2\pmod7

3^2\equiv2\pmod7

3^3\equiv2\cdot3=6\equiv-1\pmod7

(a/7)=-1,\quad a^3\equiv6\pmod7

a^{(7-1)/2}\equiv(a/7)\pmod7

b=2,\quad(2/7)=1

2^2\equiv4,\quad2^3\equiv4\cdot2\equiv1\pmod7

a=1,2,3,4,5,6

a^3\bmod7=1,1,6,1,6,6;\quad(a/7)=1,1,-1,1,-1,-1

a\equiv0\pmod7\text{ or }p=2\text{ or }p\text{ composite}

\text{outside this criterion's qualified domain}

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S09 · premise_audit · G1, S01, S06, S08 — No Jacobi symbol, reciprocity or composite square-root algorithm is inferred.
• S08 · premise_audit · G1, S02, S03, S04, S05, S07 — Even a complete table covers p=7 only, not the theorem for every odd prime.
• S07 · premise_audit · G1, S03 — The root witness 3^2=2 comes from S02, not inferred from this power result.
• S06 · premise_audit · G1, S01, S03, S05 — Retain modulo 7: integers 6 and -1 are unequal but congruent.
• S05 · premise_audit · G1, S04 — Residue 6 is not a fourth Legendre value; the signed value remains -1.
• S04 · premise_audit · G1, S01 — This calculation does not use Q7 membership and has a different role from the root table.
• S03 · premise_audit · G1, S02 — The symbol is not defined backward from the a^3 result.
• S02 · premise_audit · G1, S01 — This route independently establishes the Legendre classification.
• S01 · premise_audit · G1 — The zero class is outside the theorem domain; primality alone is insufficient.

Error controls and negative controls

Two routes

Define Legendre from all squares, then compute the half-power independently.

Sign and residue

Residue 6 modulo 7 corresponds to signed -1, not integer equality.

• The zero class modulo 7 should give +1. → reject — 0^3=0 and this criterion requires nonzero input.
• The Legendre value is integer 6. → reject — 6 represents -1 modulo 7; the symbol value is -1.
• This finite table proves the theorem for all odd primes. → reject — The table contains only p=7 and does not cover all primes.

Result and limitations

For p=7,a=-4 the Legendre value is -1 and a^3=6=-1 mod 7; 2 gives +1.

Verification and source binding

3 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: bfc2a00296ec4845bb2bac52d8d35b1658baa23430ee96f708f9268d434e17b8
• original source SHA-256: 2f805577c16cb2f7cbc2ff07f2140cb59c25f78198341aa58674efdb3cd73780
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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