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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An odd prime has $(p - 1) / 2$ nonzero quadratic residues and as many nonresidues

Statement

For an odd prime $p$ , exactly $(p - 1) / 2$ nonzero classes are quadratic residues modulo $p$ , and exactly $(p - 1) / 2$ are quadratic nonresidues. Each nonzero quadratic residue has exactly two square roots modulo $p$ , while each nonresidue has none.

Facts & Assumptions

Given: An odd prime $p$ .

[L1]

The nonzero quadratic-residue classes form the subgroup $((Z / p)^{\times})^{2} = {u^{2} : u \in (Z / p)^{\times}}$ , and this subgroup has index two in $(Z / p)^{\times}$ (The nonzero squares modulo an odd prime form an index-two subgroup).

[L2]

If $G$ is finite and $H \leq G$ , then $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L3]

Under its primitive-root, coprimality, positivity, and solubility hypotheses, $x^{m} \equiv a (m o d n)$ has exactly $\gcd (φ (n), m)$ solution classes (If $n$ has a primitive root, $\gcd (a, n) = 1$ , $m \geq 1$ , and $x^{m} \equiv a (m o d n)$ is solvable, then it has exactly $\gcd (φ (n), m)$ solution classes modulo $n$ ).

[L4]

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

[L5]

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

[L6]

The group $(Z / p)^{\times}$ is cyclic of order $p - 1$ (For every prime $p$ , the multiplicative group $(Z / p Z)^{\times}$ is cyclic).

[L7]

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

Let $Q = {u^{2} : u \in G}$ be the subgroup of nonzero square classes, which by [L1] is exactly the set of quadratic-residue classes. By [L1], $[G : Q] = 2$ for $G = (Z / p)^{\times}$ , and [L6] gives $∣ G ∣ = p - 1$ ; hence [L2] gives $∣ Q ∣ = (p - 1) / 2$ . Its complement in $G$ has the same cardinality, and since [L1] identifies $Q$ with the quadratic-residue classes, that complement consists exactly of the nonresidues.

L1L2L6given

1.2

If $a$ represents a class in $Q$ , then $p ∤ a$ , so [L7] gives $\gcd (a, p) = 1$ . Facts [L4] and [L5] discharge the prime specialisation of [L3], which gives $\gcd (p - 1, 2) = 2$ roots because $p$ is odd.

L3L4L5L7algebra

2.1

A nonresidue has no square root by the description of $Q$ in [L1], while step 1.2 gives exactly two roots for each member of $Q$ . Together with step 1.1, this proves all assertions.

L1step 1.1step 1.2∎

Depends on

Used by

Quadratic residues, roots, and the Legendre table modulo 11 Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 131 results over 22 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

N. Mascot, Legendre Symbols and Quadratic Reciprocity, Chapter 3 (standard reference, not scraped)
W. Stein, Elementary Number Theory, Section 4.1 (standard reference, not scraped)

An odd prime has (p−1)/2(p−1)/2 nonzero quadratic residues and as many nonresidues