PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-16

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Quadratic residuosity is representative-independent and the residues are the image of squaring

Statement

Let $n \geq 2$ . Whether an integer $a$ with $\gcd (a, n) = 1$ is a quadratic residue modulo $n$ depends only on its class $[a]_{n}$ . Moreover, the quadratic-residue classes are exactly

${u^{2} : u \in (Z / n)^{\times}},$

the image of squaring on the unit group.

Facts & Assumptions

Given: An integer $n \geq 2$ and integers representing unit classes modulo $n$ .

[L1]

For $\gcd (a, n) = 1$ , the integer $a$ is a quadratic residue modulo $n$ exactly when some integer $x$ satisfies $x^{2} \equiv a (m o d n)$ (Quadratic residues and nonresidues modulo an integer).

[L2]

Two classes in $Z / n$ are equal exactly when their representatives are congruent modulo $n$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ).

[L3]

The class $[a]_{n}$ is a unit exactly when $\gcd (a, n) = 1$ , and this condition depends only on the class (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L4]

A class $u \in Z / n$ is a unit exactly when some $v \in Z / n$ satisfies $u v = [1]_{n}$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L5]

Multiplication makes $Z / n$ a commutative monoid with identity $[1]_{n}$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Proof

technique · direct

1.1L1L2given

If $a \equiv b (m o d n)$ and $x^{2} \equiv a (m o d n)$ , then $[x]_{n}^{2} = [a]_{n} = [b]_{n}$ , so $x^{2} \equiv b (m o d n)$ ; reversing the roles of $a$ and $b$ gives the converse.

2.1L3L4L5step 1.1

By [L3], congruent representatives are simultaneously units. If $[x]_{n}^{2} = [a]_{n}$ and $[a]_{n}$ is a unit with inverse $v$ , then $[x]_{n} ([x]_{n} v) = [x]_{n}^{2} v = [1]_{n}$ , so $[x]_{n}$ is a unit. Thus every quadratic-residue class lies in the image of squaring on $(Z / n)^{\times}$ .

3.1L1L2L3L4L5step 1.1step 2.1∎

Conversely, let $u = [x]_{n}$ be a unit with $u^{2} = [a]_{n}$ . By [L4] there is $v$ with $u v = [1]_{n}$ , so $u^{2} v^{2} = [1]_{n}$ by [L5] and $[a]_{n} = u^{2}$ is itself a unit; [L3] then gives $\gcd (a, n) = 1$ , which is the hypothesis [L1] requires. Since $[x]_{n}^{2} = [a]_{n}$ , [L2] gives $x^{2} \equiv a (m o d n)$ , so [L1] makes $a$ a quadratic residue. Hence the quadratic-residue classes are exactly the displayed image.

Depends on

Used by

Two quadratic nonresidues modulo 15 can have a nonresidue product Counterexample
The Legendre symbol is well defined on residue classes Proposition
The nonzero squares modulo an odd prime form an index-two subgroup Theorem

Dependency tree · two levels

24 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)
W. Stein, Elementary Number Theory, Section 4.1 (standard reference, not scraped)