CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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A soluble square congruence need not define a quadratic residue

Statement refuted

Solubility of $x^{2} \equiv a (m o d p)$ does not by itself make $a$ a quadratic residue. For every odd prime $p$ , the target $a = 0$ gives a soluble congruence but belongs to neither the residue nor the nonresidue class.

Facts & Assumptions

Given: An odd prime $p$ and the target $a = 0$ .

[L1]

For every integer $a$ and odd prime $p$ , $x^{2} \equiv a (m o d p)$ has exactly $1 + (a / p)$ solution classes ( $x^{2} \equiv a (m o d p)$ has exactly $1 + (a / p)$ solution classes).

[L2]

The terms quadratic residue and quadratic nonresidue modulo $n$ are defined only for integers coprime to $n$ (Quadratic residues and nonresidues modulo an integer).

[L3]

The Legendre symbol has value $0$ when its numerator is divisible by the odd prime denominator (The Legendre symbol, including its zero value).

Counterexample

technique · direct

1.1L1L3given

Since $(0 / p) = 0$ by [L3], fact [L1] gives exactly one solution class to $x^{2} \equiv 0 (m o d p)$ ; it is visibly the zero class.

2.1L2L3step 1.1∎

Yet $\gcd (0, p) = p \neq 1$ , so [L2] puts zero outside the quadratic residue and nonresidue dichotomy, and [L3] assigns it the separate Legendre value zero. Thus the congruence is soluble without making its target a quadratic residue.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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)