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CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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Odd-prime tests alone miss the obstruction to 13 being a square modulo 24

Statement refuted

For an even modulus, testing whether a unit is a square modulo every odd prime divisor does not suffice to prove that it is a square modulo the original modulus.

Facts & Assumptions

Given: The unit 13 modulo 24=8⋅3.

[L1]

For k≥3, an odd integer is a square modulo 2k if and only if it is congruent to 1 modulo 8 (Unit square criterion and root count modulo powers of two).

[L2]

A unit square modulo n is a square modulo every prime-power factor of n (A unit is a square modulo n exactly when it is a square at every prime-power factor).

Counterexample

technique · direct
1.1L1givenalgebra

The only odd prime divisor of 24 is 3, and 13≡1(mod3), so it passes the odd-prime square test. But 13≡5(mod8), and [L1] shows that it is not a square modulo 8.

2.1step 1.1L2∎

If 13 were a square modulo 24, the forward implication in [L2] would make it a square modulo the prime-power factor 8, contradicting step 1.1. Hence the odd-prime test misses the two-part obstruction.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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.