CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Primitive sums of two squares are closed under products unless both factors are even

Statement

Let $m, n$ be positive integers that have primitive two-square representations. If $m$ and $n$ are not both even, then $m n$ has a primitive two-square representation.

Facts & Assumptions

Given: Positive primitively represented integers $m, n$ .

[L1]

A positive integer $n$ has a primitive two-square representation if and only if $v_{2} (n) \leq 1$ and no prime $q \equiv 3 (m o d 4)$ divides $n$ (Characterisation of primitive sums of two squares).

[L2]

For nonzero integers $a, b$ , $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ ( $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero).

[L3]

For a prime $p$ and a nonzero integer $a$ , $v_{p} (a) \geq 1$ if and only if $p ∣ a$ (For a prime $p$ and a nonzero integer $a$ : $p^{v_{p} (a)} ∣ a$ and $p^{v_{p} (a) + 1} ∤ a$ ; $p^{k} ∣ a$ holds exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = v_{p} (- 1) = 0$ ; and $v_{p} (p) = 1$ ).

Proof

technique · direct

1.1

By [L1] and [L3], every prime congruent to three modulo four has valuation zero in each factor. By [L2], its valuation in the product is zero, so [L3] shows that it does not divide the product.

givenL1L2L3

1.2

Again by [L1]–[L3], $v_{2} (m n) = v_{2} (m) + v_{2} (n) \leq 1$ because at least one of $m, n$ is odd and therefore has $2$ -adic valuation zero.

givenL1L2L3algebra

2.1

The two conditions in [L1] hold for $m n$ , so the product has a primitive two-square representation.

step 1.1step 1.2L1∎

Depends on

Used by

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Sources

P. Hackman, Elementary Number Theory, Chapter E, Corollary E.II.8(a) (standard reference, not scraped)