False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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FALSE: a prime has one four-square representation up to order and signs

Statement

False claim: for every prime $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), any two representations of $p$ as a sum of four integer squares (Representations as sums of four squares) are equivalent up to signs and order; that is, a prime has exactly one four-square representation up to permuting the coordinates and changing their signs.

Facts & Assumptions

Given: The integer $31$ and the quadruples $(5, 2, 1, 1)$ and $(3, 3, 3, 2)$ .

[A1]

The false claim: for every prime $p$ , any two representations of $p$ as a sum of four integer squares are equivalent up to signs and order.

[F1]

A representation of a nonnegative integer $n$ as a sum of four squares is an ordered quadruple $(a, b, c, d) \in Z^{4}$ with $n = a^{2} + b^{2} + c^{2} + d^{2}$ ; two representations are equivalent up to signs and order exactly when their multisets of absolute values coincide (Representations as sums of four squares).

[F2]

An integer $p$ is prime when $p > 1$ and $d ∣ p$ with $d > 0$ force $d = 1$ or $d = p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

Refutation

technique · direct

1.1

The integer $31$ is prime: $31 > 1$ , and if $31 = d e$ with integers $1 < d \leq e$ then $d^{2} \leq d e = 31$ , so $d \leq 5$ ; but $31 = 2 \cdot 15 + 1 = 3 \cdot 10 + 1 = 4 \cdot 7 + 3 = 5 \cdot 6 + 1$ , so none of $2, 3, 4, 5$ divides $31$ , and by [F2] its only positive divisors are $1$ and $31$ .

givenF2algebra

1.2

Both displays are representations of $31$ : $5^{2} + 2^{2} + 1^{2} + 1^{2} = 25 + 4 + 1 + 1 = 31$ and $3^{2} + 3^{2} + 3^{2} + 2^{2} = 9 + 9 + 9 + 4 = 31$ .

givenF1algebra

2.1

The multisets of absolute values are ${5, 2, 1, 1}$ and ${3, 3, 3, 2}$ , which differ — the first contains $5$ and the second does not — so by the criterion in [F1] the two representations of step 1.2 are not equivalent up to signs and order; with step 1.1 this contradicts [A1] at $p = 31$ , and the claim is false.

step 1.1step 1.2A1F1algebra∎

Remarks

Where the claim comes from. For two squares the corresponding statement is true: A prime congruent to $1$ modulo $4$ has one two-square representation up to signs and order says that a prime congruent to $1$ modulo $4$ has one representation as a sum of two squares up to signs and order. The claim refuted above is that statement with two coordinates replaced by four.

What survives of the analogy. Nothing of the two-square proof transfers. It turns on such a prime having a primitive two-square representation whose factorisation is controlled, and with four coordinates the extra room admits genuinely different multisets of absolute values; $31$ is one witness and not a special one.

Depends on

Used by

Nothing in the library uses this result yet.

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

Keith Conrad, Proofs by Descent, §6, Example 6.1 (standard reference, not scraped)