TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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

1.1

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.

L1given

2.1

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

L2L5step 1.1algebra

3.1

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 $a^{m} \equiv (- 1)^{N (a, p)} (m o d p)$ .

L3L5L6step 2.1

4.1

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 $Z$ .

L4L7step 3.1givenalgebra∎

Depends on

Used by

The signed half-system for 3 modulo 11 gives (3/11)=1 Example
Second supplement: (2/p)=(-1)^(p²-1)/8 Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 106 results over 28 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)