TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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The Legendre symbol is multiplicative for all integer numerators

Statement

For every odd prime $p$ and all integers $a, b$ ,

$(\frac{a b}{p}) = (\frac{a}{p}) (\frac{b}{p}) .$

Consequently, if $p ∤ c$ , then $(a c^{2} / p) = (a / p)$ .

Facts & Assumptions

Given: An odd prime $p$ and integers $a, b, c$ .

[L1]

The Legendre symbol is $0$ when its numerator is divisible by $p$ , $1$ on a quadratic residue, and $- 1$ on a quadratic nonresidue (The Legendre symbol, including its zero value).

[L2]

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

[L3]

Restricted to $(Z / p)^{\times}$ , the Legendre symbol is a homomorphism to ${\pm 1}$ whose kernel is the nonzero square subgroup (On the units, the Legendre symbol is the unique nontrivial homomorphism to ${\pm 1}$ ).

Proof

technique · direct

1.1L1L2given

If $p$ divides $a$ or $b$ , then it divides $a b$ . By [L1], the left side is zero and one factor on the right is zero, so the identity holds.

2.1L2L3step 1.1

If $p$ divides neither factor, then [L2] makes $[a]_{p}$ and $[b]_{p}$ units. The homomorphism identity in [L3] gives the displayed multiplicativity.

3.1L1L3step 2.1∎

If $p ∤ c$ , then $[c]_{p}^{2}$ lies in the kernel described by [L3], so $(c^{2} / p) = 1$ . Applying the proved multiplicative identity to $a$ and $c^{2}$ gives $(a c^{2} / p) = (a / p)$ .

Depends on

Used by

Dependency tree · two levels

16 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.I (standard reference, not scraped)
A. Gorodnik, Number Theory, Lecture 9, Theorem 1.6 (standard reference, not scraped)