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Sums of four squares are closed under products

Statement

Let $m$ and $n$ be nonnegative integers. If each of $m$ and $n$ is a sum of four integer squares (Representations as sums of four squares), then $m n$ is a sum of four integer squares.

Facts & Assumptions

Given: Nonnegative integers $m$ and $n$ , each a sum of four integer squares.

[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 $n = a^{2} + b^{2} + c^{2} + d^{2}$ ; the integer is a sum of four integer squares when at least one such quadruple exists (Representations as sums of four squares).

[L1]

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).

Proof

technique · direct

1.1

Fix quadruples $(x_{1}, x_{2}, x_{3}, x_{4})$ and $(y_{1}, y_{2}, y_{3}, y_{4})$ of integers with $m = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}$ and $n = y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2}$ , which the hypothesis supplies.

givenF1choose

2.1

With $z_{1}, z_{2}, z_{3}, z_{4}$ formed from those eight integers by the displayed formulas, $m n = (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2}) = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ .

step 1.1L1

3.1

The four integers $z_{1}, z_{2}, z_{3}, z_{4}$ therefore form a quadruple in $Z^{4}$ whose squares sum to $m n$ , so $m n$ is a sum of four integer squares; no coordinate is required to be positive or nonzero, so the argument is unaffected when a $z_{i}$ vanishes or is negative, and it covers $m = 0$ and $n = 0$ , whose quadruple $(0, 0, 0, 0)$ satisfies the hypothesis and returns $z_{1} = z_{2} = z_{3} = z_{4} = 0$ .

step 2.1F1algebra∎

Remarks

What the identity does and does not give. The four coordinates $z_{i}$ are determined by the two chosen quadruples, so a different choice of representation of $m$ or of $n$ generally produces a different representation of $m n$ . The statement asserts existence only; it makes no claim about how many representations $m n$ has, nor that every representation of $m n$ arises this way.

Why the closure is needed. Reducing Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares to Every prime is a sum of four integer squares requires exactly this: a factorisation of $n$ into primes is useless unless the property being proved is inherited by products.

Depends on

Used by

Dependency tree · one level

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Sources

Keith Conrad, Proofs by Descent, §6, Lemma 6.2 (standard reference, not scraped)
Evan Dummit, Number Theory (part 9): The Geometry of Numbers, §9.1.2, Lemma 1 (standard reference, not scraped)