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

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Building a representation of $105$ from its prime factors

Example

Factor $105 = 3 \cdot 5 \cdot 7$ and take the prime representations

$3 = 1^{2} + 1^{2} + 1^{2} + 0^{2}, 5 = 2^{2} + 1^{2} + 0^{2} + 0^{2}, 7 = 2^{2} + 1^{2} + 1^{2} + 1^{2} .$

Multiplying the first two with the identity of Euler's four-square product identity, applied to $(1, 1, 1, 0)$ and $(2, 1, 0, 0)$ , gives $(3, - 1, - 2, 1)$ and

$15 = 3^{2} + (- 1)^{2} + (- 2)^{2} + 1^{2} .$

Multiplying that result by the third, applied to $(3, - 1, - 2, 1)$ and $(2, 1, 1, 1)$ , gives $(4, 8, 5, 0)$ and

$105 = 4^{2} + 8^{2} + 5^{2} + 0^{2} .$

This is the route the proof of Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares takes: prime factors first, then closure under products.

Facts & Assumptions

Given: The factorisation $105 = 3 \cdot 5 \cdot 7$ and the three displayed prime representations.

[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 $n = a^{2} + b^{2} + c^{2} + d^{2}$ (Representations as sums of four squares).

[L1]

Let $m$ and $n$ be nonnegative integers; if each of $m$ and $n$ is a sum of four integer squares, then $m n$ is a sum of four integer squares (Sums of four squares are closed under products).

[L2]

For all integers $x_{1}, \dots, x_{4}, y_{1}, \dots, y_{4}$ , setting $z_{1} = x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + x_{4} y_{4}$ , $z_{2} = x_{1} y_{2} - x_{2} y_{1} - x_{3} y_{4} + x_{4} y_{3}$ , $z_{3} = x_{1} y_{3} + x_{2} y_{4} - x_{3} y_{1} - x_{4} y_{2}$ and $z_{4} = x_{1} y_{4} - x_{2} y_{3} + x_{3} y_{2} - x_{4} y_{1}$ gives $(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2}) = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ (Euler's four-square product identity).

Verification

technique · direct

1.1

The three displays are representations: $1 + 1 + 1 + 0 = 3$ , $4 + 1 + 0 + 0 = 5$ and $4 + 1 + 1 + 1 = 7$ ; also $3 \cdot 5 = 15$ and $15 \cdot 7 = 105$ , so $3 \cdot 5 \cdot 7 = 105$ .

givenF1algebra

2.1

Applying [L2] to $x = (1, 1, 1, 0)$ and $y = (2, 1, 0, 0)$ gives $z_{1} = 1 \cdot 2 + 1 \cdot 1 + 1 \cdot 0 + 0 \cdot 0 = 3$ , $z_{2} = 1 \cdot 1 - 1 \cdot 2 - 1 \cdot 0 + 0 \cdot 0 = - 1$ , $z_{3} = 1 \cdot 0 + 1 \cdot 0 - 1 \cdot 2 - 0 \cdot 1 = - 2$ and $z_{4} = 1 \cdot 0 - 1 \cdot 0 + 1 \cdot 1 - 0 \cdot 2 = 1$ , whose squares sum to $9 + 1 + 4 + 1 = 15 = 3 \cdot 5$ .

step 1.1L2algebra

3.1

Applying [L2] to $x = (3, - 1, - 2, 1)$ and $y = (2, 1, 1, 1)$ gives $z_{1} = 3 \cdot 2 + (- 1) \cdot 1 + (- 2) \cdot 1 + 1 \cdot 1 = 4$ , $z_{2} = 3 \cdot 1 - (- 1) \cdot 2 - (- 2) \cdot 1 + 1 \cdot 1 = 8$ , $z_{3} = 3 \cdot 1 + (- 1) \cdot 1 - (- 2) \cdot 2 - 1 \cdot 1 = 5$ and $z_{4} = 3 \cdot 1 - (- 1) \cdot 1 + (- 2) \cdot 1 - 1 \cdot 2 = 0$ , whose squares sum to $16 + 64 + 25 + 0 = 105 = 15 \cdot 7$ .

step 2.1L2algebra

4.1

So $(4, 8, 5, 0)$ is a representation of $105$ in the sense of [F1]; the two multiplications are the two instances of [L1] that the factorisation calls for, and the identity has supplied an explicit quadruple at each.

step 1.1step 2.1step 3.1L1F1∎

Remarks

Negative intermediate coordinates are harmless. The quadruple produced at the first multiplication has two negative entries, and they are carried into the second multiplication unchanged. Only squares are read off at the end, and the identity of Euler's four-square product identity holds for negative inputs as it does for positive ones.

The order of multiplication is a choice. Multiplying $5$ and $7$ first, or $3$ and $7$ first, produces different quadruples for the same $105$ ; the construction is not canonical, and Sums of four squares are closed under products claims existence rather than a preferred witness.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · one level

3 results within one dependency step 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, Lemma 6.2 (standard reference, not scraped)

Building a representation of 105105 from its prime factors