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Powers of primes congruent to $1$ modulo $4$ have primitive two-square representations

Statement

Every natural power of a prime congruent to $1$ modulo $4$ has a primitive two-square representation.

Facts & Assumptions

Given: A prime $p \equiv 1 (m o d 4)$ and an exponent $e \in N$ .

[F1]

A two-square representation is primitive when its coordinate gcd is $1$ (Representations and primitive representations as sums of two squares).

[L1]

For all integers $a, b, c, d$ , $(a^{2} + b^{2}) (c^{2} + d^{2}) = (a c - b d)^{2} + (a d + b c)^{2} = (a c + b d)^{2} + (a d - b c)^{2}$ (The Brahmagupta–Fibonacci two-square identity).

[L2]

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).

[L3]

Every integer greater than $1$ has a prime divisor (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime).

[L4]

If a prime $ℓ$ divides $u v$ , then $ℓ ∣ u$ or $ℓ ∣ v$ (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ ).

[L5]

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

[L6]

If a positive integer is written as a finite product of powers of distinct primes, every exponent equals the corresponding canonical valuation (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).

Proof

technique · induction

1.1

The pair $(1, 0)$ primitively represents $p^{0} = 1$ .

F1basealgebra

1.2

By [L2], choose $a, b$ with $p = a^{2} + b^{2}$ . Neither coordinate is zero, and divisibility of either coordinate by $p$ would force divisibility of the other and then $p^{2} ∣ p$ ; in particular $p$ divides neither coordinate. Any common prime divisor would have square dividing $p$ , so $(a, b)$ is primitive; the coordinates have opposite parity because $p$ is odd.

givenL2F1choosealgebra

2.1

Assume $p^{e} = x^{2} + y^{2}$ primitively. The two sign variants in [L1] give representations of $p^{e + 1}$ with coordinate pairs $(a x - b y, a y + b x)$ and $(a x + b y, a y - b x)$ .

ihstep 1.2F1L1construct

3.1

If a prime $ℓ$ divides both coordinates of either candidate, then $ℓ^{2} ∣ p^{e + 1}$ by step 2.1. The uniqueness of prime exponents in [L6], or [L4] iterated through the power, forces $ℓ = p$ .

step 2.1L3L4L6algebra

3.2

If both candidates were coordinatewise divisible by $p$ , their sums and differences would show that $p$ divides $2 a x, 2 a y, 2 b x, 2 b y$ . Since $p$ is odd and neither $a$ nor $b$ is divisible by $p$ , [L4] would give $p ∣ x$ and $p ∣ y$ , contradicting the induction hypothesis.

step 2.1step 1.2L4ihalgebra

4.1

Thus at least one candidate has no common prime divisor; by [L3] its coordinate gcd cannot exceed one, so it is primitive. Step 1.1 and [L5] complete the induction.

step 1.1step 3.1step 3.2L3F1L5discharge-induction∎

Depends on

Used by

Characterisation of primitive sums of two squares Theorem

Dependency tree · two levels

46 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, Lemma E.II.6 (standard reference, not scraped)

Powers of primes congruent to 11 modulo 44 have primitive two-square representations