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

Statement

For every integer $a$ and odd prime $p$ ,

$(\frac{a}{p}) \equiv a^{(p - 1) / 2} (m o d p) .$

Facts & Assumptions

Given: An integer $a$ and an odd prime $p$ .

[L1]

The Legendre symbol is $0$ when $p ∣ a$ , $1$ when $a$ is a quadratic residue modulo $p$ , and $- 1$ when $a$ is a quadratic nonresidue (The Legendre symbol, including its zero value).

[L2]

If $n \geq 1$ admits a primitive root, $\gcd (a, n) = 1$ , $m \geq 1$ , and $d = \gcd (m, φ (n))$ , then $x^{m} \equiv a (m o d n)$ is soluble exactly when $a^{φ (n) / d} \equiv 1 (m o d n)$ (Euler's criterion: if $n$ has a primitive root, $\gcd (a, n) = 1$ , and $m \geq 1$ , then $x^{m} \equiv a (m o d n)$ is solvable if and only if $a^{φ (n) / \gcd (φ (n), m)} \equiv 1 (m o d n)$ ).

[L3]

Every prime admits a primitive root modulo that prime (Every prime modulus admits a primitive root).

[L4]

For every prime $p$ , $φ (p) = p - 1$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ).

[L5]

If $p ∤ a$ , then $a^{p - 1} \equiv 1 (m o d p)$ (Fermat's little theorem: for prime $p$ , $p ∤ a$ implies $a^{p - 1} \equiv 1 (m o d p)$ , and always $a^{p} \equiv a (m o d p)$ ).

[L6]

For every prime $p$ , the quotient $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L7]

Every field is an integral domain (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring).

[L8]

A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L9]

If $p ∤ a$ , then $p$ and $a$ are coprime (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).

Proof

technique · direct

1.1

If $p ∣ a$ , then [L1] gives $(a / p) = 0$ , while $(p - 1) / 2 \geq 1$ and hence $a^{(p - 1) / 2} \equiv 0 (m o d p)$ . The criterion holds in this case.

L1givenalgebra

1.2

Suppose $p ∤ a$ . By [L9], $\gcd (a, p) = 1$ ; [L3] and [L4] specialise [L2] with $m = 2$ and $d = 2$ . Thus $a$ is a square modulo $p$ exactly when $a^{(p - 1) / 2} \equiv 1 (m o d p)$ .

L1L2L3L4L9algebra

1.3

Put $y = [a^{(p - 1) / 2}]_{p}$ . By [L5], $y^{2} = [1]_{p}$ . Facts [L6] and [L7] make $Z / p$ an integral domain, and [L8] says that the degree-two polynomial $X^{2} - 1$ has at most two roots there. The two distinct roots $[1]_{p}$ and $[- 1]_{p}$ already exist because $p$ is odd, so $y$ is one of them.

L5L6L7L8algebra

2.1

In the unit case, step 1.2 gives $y = [1]_{p}$ exactly for a quadratic residue; otherwise step 1.3 forces $y = [- 1]_{p}$ . These are precisely the two values prescribed by [L1]. Combining this with step 1.1 proves the congruence for every integer $a$ .

L1step 1.1step 1.2step 1.3∎

Depends on

Used by

Euler's criterion computes (3/726377359) by repeated squaring Example
First supplement: (-1/p)=(-1)^(p-1)/2 Theorem
Gauss's quadratic-residue lemma Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 117 results over 26 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.I (standard reference, not scraped)
W. Stein, Elementary Number Theory, Section 4.2 (standard reference, not scraped)
A. Gorodnik, Number Theory, Lecture 9, Theorem 1.5 (standard reference, not scraped)

Euler's criterion: (a/p)≡a(p−1)/2(modp)(a/p)≡a(p−1)/2(modp)