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Prime powers represented as sums of two squares

Statement

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 a prime $q \equiv 3 (m o d 4)$ is representable exactly when its exponent is even. The exponent ranges over all of $N$ , including zero.

Facts & Assumptions

Given: A prime and a natural exponent.

[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]

A prime $p$ is a sum of two integer squares if and only if $p = 2$ or $p \equiv 1 (m o d 4)$ (Fermat's two-square theorem for primes).

[L4]

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

Proof

technique · direct

1.1F1basealgebra

For every prime $ℓ$ , the exponent-zero power is $ℓ^{0} = 1 = 1^{2} + 0^{2}$ .

1.2F1constructalgebra

If $q \equiv 3 (m o d 4)$ , every even power has the explicit representation $q^{2 j} = (q^{j})^{2} + 0^{2}$ .

1.3F1L2L4algebradischarge-induction

If $q^{2 j + 1} = x^{2} + y^{2}$ , then [L2] gives $x = q x_{1}$ and $y = q y_{1}$ , so $q^{2 j - 1} = x_{1}^{2} + y_{1}^{2}$ when $j \geq 1$ ; at $j = 0$ , the same divisibility would make $q^{2}$ divide $q$ , which is impossible. Induction on $j$ repeatedly reduces any alleged odd-exponent representation to that impossible base case.

2.1step 1.1L1L3L4discharge-induction

The prime $2$ and every prime $p \equiv 1 (m o d 4)$ are represented by [L3]. Multiplying an induction-stage representation by the prime representation and using [L1] gives one for the next power, so [L4] represents all their natural powers.

3.1step 1.1step 2.1step 1.2step 1.3∎

Steps 1.1 and 2.1 handle $2$ and primes congruent to one modulo four, while steps 1.2 and 1.3 prove both directions for primes congruent to three modulo four.

Depends on

Used by

Characterisation of positive integers that are sums of two squares Theorem

Dependency tree · two levels

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