CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-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.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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$3$ and $5$ are sums of three squares and $15$ is not

Statement refuted

False claim: if two nonnegative integers are each a sum of three integer squares, then so is their product.

The pair $3$ , $5$ refutes it: $3 = 1^{2} + 1^{2} + 1^{2}$ and $5 = 2^{2} + 1^{2} + 0^{2}$ , while $15 = 3 \cdot 5$ satisfies $15 \equiv 7 (m o d 8)$ and is therefore excluded by Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ are not sums of three integer squares.

Facts & Assumptions

Given: The integers $3$ , $5$ and $15$ .

[A1]

The false claim: if $m$ and $n$ are nonnegative integers, each a sum of three integer squares, then $m n$ is a sum of three integer squares.

[F1]

For $a, b, n \in Z$ , $a \equiv b (m o d n)$ means $n ∣ (a - b)$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L1]

For $a \in N$ and a positive integer $m$ with $m \equiv 7 (m o d 8)$ , there are no integers $x, y, z$ with $4^{a} m = x^{2} + y^{2} + z^{2}$ (Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ are not sums of three integer squares).

Counterexample

technique · direct

1.1

Both factors are sums of three integer squares: $1^{2} + 1^{2} + 1^{2} = 3$ and $2^{2} + 1^{2} + 0^{2} = 5$ .

givenalgebra

1.2

The product is $3 \cdot 5 = 15$ , and $15 - 7 = 8$ , so $15 \equiv 7 (m o d 8)$ by [F1]; taking $a = 0$ and $m = 15$ , which is a positive integer congruent to $7$ modulo $8$ , [L1] gives that no integers $x, y, z$ satisfy $4^{0} \cdot 15 = 15 = x^{2} + y^{2} + z^{2}$ .

givenL1F1algebra

2.1

So $3$ and $5$ are sums of three integer squares while $15$ is not, and [A1] is false.

step 1.1step 1.2A1∎

Remarks

The failure is not universal. Some products of three-square integers are again sums of three squares: $3 \cdot 3 = 9 = 3^{2} + 0^{2} + 0^{2}$ is one. The claim refuted above is that this always happens, and a single product for which it fails is what settles it.

What this separates. Sums of two squares are closed under products by the Brahmagupta–Fibonacci identity (Sums of two squares are closed under products), and sums of four squares by Sums of four squares are closed under products. Three squares admit no such product identity, and this pair is why: an identity expressing $(x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2})$ as a sum of three squares of integer bilinear forms would make $15$ a sum of three squares.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Evan Dummit, Number Theory (part 9): The Geometry of Numbers, §9.1.3 (standard reference, not scraped)

33 and 55 are sums of three squares and 1515 is not