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Characterisation of positive integers that are sums of two squares

Statement

A positive integer $n$ is a sum of two squares if and only if every prime $q \equiv 3 (m o d 4)$ occurs to an even exponent in its canonical prime factorisation.

Facts & Assumptions

Given: A positive integer $n$ .

[F1]

A representation of a nonnegative integer $n$ as a sum of two squares is an ordered pair $(x, y) \in Z^{2}$ such that $n = x^{2} + y^{2}$ (Representations and primitive representations as sums of two squares).

[L1]

The product of two nonnegative integers representable as sums of two squares is again representable as a sum of two squares (Sums of two squares are closed under products).

[L2]

If $q \equiv 3 (m o d 4)$ is prime and $q ∣ x^{2} + y^{2}$ , then $q ∣ x$ and $q ∣ y$ (A prime congruent to $3$ modulo $4$ divides both coordinates of a divisible two-square sum).

[L3]

Every power of $2$ and every power of a prime $p \equiv 1 (m o d 4)$ is a sum of two squares; a power of $q \equiv 3 (m o d 4)$ is representable exactly when its exponent is even (Prime powers represented as sums of two squares).

[L4]

A positive integer is the finite product of the powers of its prime divisors with their canonical valuations (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

[L5]

For a nonzero integer, $p^{k}$ divides it if and only if $k \leq v_{p}$ (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 $v_{p} (p) = 1$ ).

[L6]

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).

[L7]

If a property holds at $0$ and passes from $j$ to $j + 1$ , it holds for every $j \in N$ (The principle of mathematical induction).

Proof

technique · direct

1.1

For the reverse direction at $n = 1$ , the canonical factorisation is empty and $(1, 0)$ represents $1$ .

givenF1L4algebra

1.2

For the reverse direction at $n > 1$ , assume every three-mod-four prime has even valuation. By [L3], every prime-power factor in [L4] is representable.

givenL3L4

1.3

For the forward direction, suppose $n = x^{2} + y^{2}$ and let $q \equiv 3 (m o d 4)$ divide $n$ . By [L2], $x = q x_{1}$ and $y = q y_{1}$ , so $n = q^{2} n_{1}$ with $n_{1} = x_{1}^{2} + y_{1}^{2}$ .

givenF1L2algebra

2.1

Repeatedly applying [L1] to the finite list of factors from step 1.2 gives a representation of their product $n$ ; step 1.1 supplies the empty-list case.

step 1.1step 1.2L1F1algebra

2.2

By [L5] and [L6], step 1.3 lowers the $q$ -adic valuation by exactly two: $v_{q} (n_{1}) = v_{q} (n) - 2$ .

step 1.3L5L6algebra

3.1

If $v_{q} (n)$ were odd, induction on the number of two-step reductions in steps 1.3 and 2.2 would eventually give a represented integer of $q$ -valuation one. Applying step 1.3 once more would make its valuation at least two, a contradiction. Thus $v_{q} (n)$ is even.

step 1.3step 2.2L7discharge-induction

4.1

Steps 2.1 and 3.1 prove the reverse and forward directions, respectively.

step 2.1step 3.1∎

Depends on

Used by

Squarefree sums of two squares Corollary
539=7²·11 is not a sum of two squares Counterexample

Dependency tree · two levels

44 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

P. Hackman, Elementary Number Theory, Chapter E, Theorem E.II.2 (standard reference, not scraped)
W. Stein, Elementary Number Theory: Primes, Congruences, and Secrets, Theorem 5.7.1 (standard reference, not scraped)