TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Gauss's quadratic-residue lemma

Statement

Let $p$ be an odd prime and let $p ∤ a$ . Let $N (a, p)$ be the number of least positive residues of

$a, 2 a, \dots, \frac{p - 1}{2} a$

modulo $p$ that exceed $p / 2$ . Then

$(\frac{a}{p}) = (- 1)^{N (a, p)} .$

Facts & Assumptions

Given: An odd prime $p$ , an integer $a$ with $p ∤ a$ , and $m = (p - 1) / 2$ .

[L1]

There are unique signs $ε_{j} \in {\pm 1}$ and a permutation $r_{1}, \dots, r_{m}$ of $1, \dots, m$ with $a j \equiv ε_{j} r_{j} (m o d p)$ for $1 \leq j \leq m$ (Multiplication by $a$ with $p ∤ a$ permutes an odd prime's signed half-system up to sign).

[L2]

A finite indexed family in a monoid has a recursively defined product (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L3]

A class $[u]_{p}$ is a unit exactly when $\gcd (u, p) = 1$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L4]

Euler's criterion gives $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ (Euler's criterion: $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ ).

[L5]

Multiplication on $Z / p$ is associative and commutative with identity (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L6]

If $p$ is prime and $p ∤ a$ , then $\gcd (a, p) = 1$ (For a prime $p$ and any integer $a$ , $\gcd (p, a)$ is $p$ when $p ∣ a$ and $1$ otherwise; so $p ∤ a$ makes $p$ and $a$ coprime).

[L7]

For an odd prime $p$ , $(\frac{a}{p}) = 1$ when $p ∤ a$ and $a$ is a quadratic residue modulo $p$ , and $(\frac{a}{p}) = - 1$ when $p ∤ a$ and $a$ is a quadratic nonresidue modulo $p$ (The Legendre symbol, including its zero value).

Proof

technique · direct

1.1L1given

Use [L1] to write $a j \equiv ε_{j} r_{j} (m o d p)$ for $1 \leq j \leq m$ . A sign is negative exactly when the least positive residue of $a j$ exceeds $p / 2$ , so exactly $N (a, p)$ of the signs are negative.

2.1L2L5step 1.1algebra

Multiply the $m$ congruences using [L2] and [L5]. Since the $r_{j}$ permute $1, \dots, m$ , this gives $a^{m} m! \equiv (- 1)^{N (a, p)} m! (m o d p)$ .

3.1L3L5L6step 2.1

Every factor $j$ satisfies $1 \leq j \leq m < p$ , so $p ∤ j$ and [L6] gives $\gcd (j, p) = 1$ ; then [L3] makes each $[j]_{p}$ a unit, so their product $[m!]_{p}$ is a unit and can be cancelled from step 2.1. Hence $a^{m} \equiv (- 1)^{N (a, p)} (m o d p)$ .

4.1L4L7step 3.1givenalgebra∎

By [L4], $(a / p) \equiv a^{m} \equiv (- 1)^{N (a, p)} (m o d p)$ . Since $p ∤ a$ , [L7] gives $(a / p) \in {1, - 1}$ , and $(- 1)^{N (a, p)}$ is likewise $1$ or $- 1$ ; two such integers differing by a multiple of the odd prime $p$ differ by at most $2 < p$ , so the congruence is equality in $Z$ .

Depends on

Used by

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

H. Hackman, Elementary Number Theory, Chapter D, Section D.IV (standard reference, not scraped)
W. Stein, Elementary Number Theory, Lemma 4.3.1 (standard reference, not scraped)