TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Fermat's two-square theorem for primes

Statement

A prime $p$ is a sum of two integer squares if and only if $p = 2$ or $p \equiv 1 (m o d 4)$ (Representations and primitive representations as sums of two squares).

Facts & Assumptions

Given: A prime $p$ .

[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]

If $p$ is prime, $p ∤ a$ , then there are nonzero integers $r, s$ with $∣ r ∣, ∣ s ∣ < \sqrt{p}$ and $r a \equiv s (m o d p)$ (Thue's lemma on small nonzero representatives).

[L2]

For an odd prime $p$ , $(- 1 / p) = 1$ if and only if $p \equiv 1 (m o d 4)$ (First supplement: $(- 1 / p) = (- 1)^{(p - 1) / 2}$ ).

[F2]

For an odd prime $p$ , $(a / p) = 1$ means that $p ∤ a$ and $a$ is a quadratic residue modulo $p$ (The Legendre symbol, including its zero value).

[F3]

The congruence $a \equiv b (m o d n)$ means that $n ∣ (a - b)$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

Proof

technique · direct

1.1

If an odd prime satisfies $p = x^{2} + y^{2}$ , the square residues modulo $4$ show that $x, y$ have opposite parity and hence $p \equiv 1 (m o d 4)$ .

givenF1algebra

1.2

The remaining even prime has the representation $2 = 1^{2} + 1^{2}$ .

F1algebra

1.3

For the converse direction, suppose $p \equiv 1 (m o d 4)$ . Then $p$ is odd, and [L2] and [F2] provide an integer $a$ with $a^{2} \equiv - 1 (m o d p)$ and $p ∤ a$ .

givenL2F2F3choose

2.1

Apply [L1] to this $a$ to obtain nonzero integers $r, s$ with $∣ r ∣, ∣ s ∣ < \sqrt{p}$ and $r a \equiv s (m o d p)$ .

step 1.3L1F3

3.1

Squaring the congruence in step 2.1 and using $a^{2} \equiv - 1$ gives $p ∣ r^{2} + s^{2}$ . Moreover $0 < r^{2} + s^{2} < p + p = 2 p$ . The only positive multiple of $p$ below $2 p$ is $p$ , so $p = r^{2} + s^{2}$ .

step 1.3step 2.1F1F3algebra

4.1

Step 1.1 proves necessity for odd primes, step 1.2 handles $p = 2$ , and step 3.1 proves sufficiency when $p \equiv 1 (m o d 4)$ .

step 1.1step 1.2step 3.1∎

Depends on

Used by

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