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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every nonzero residue modulo an odd prime is a sum of two squares

Statement

Let $p$ be an odd prime and let $a \in Z$ with $p ∤ a$ . Then there are integers $x, y$ such that

$x^{2} + y^{2} \equiv a (m o d p) .$

Facts & Assumptions

Given: An odd prime $p$ and a nonzero class $[a] \in Z / p$ .

[L1]

For an odd prime $p$ , exactly $(p - 1) / 2$ nonzero classes are quadratic residues modulo $p$ (An odd prime has $(p - 1) / 2$ nonzero quadratic residues and as many nonresidues).

[L2]

The quotient $Z / p$ has $p$ elements (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L3]

If finite sets $A$ and $B$ are disjoint, then $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

[L4]

A subset of a finite set is finite and has cardinality at most that of the ambient set (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[L5]

For every $n \in N$ , $(Z / n, +, [0]_{n})$ is an abelian group, and multiplication distributes over addition on both sides (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Proof

technique · contradiction

1.1

Let $Q$ be the set of all square classes in $Z / p$ , including zero. By [L1], $∣ Q ∣ = (p + 1) / 2$ .

givenL1algebra

2.1

Translation and negation are bijections of the additive group, so $a - Q : = {a - z : z \in Q}$ also has $(p + 1) / 2$ elements.

step 1.1L5algebraconstruct

3.1

Suppose, for contradiction, that $Q$ and $a - Q$ are disjoint. Then [L3] gives $∣ Q \cup (a - Q) ∣ = (p + 1) / 2 + (p + 1) / 2 = p + 1$ .

step 1.1step 2.1L3assume-contraalgebra

4.1

But $Q \cup (a - Q) \subseteq Z / p$ , so [L2] and [L4] give $∣ Q \cup (a - Q) ∣ \leq p$ , contradicting step 3.1.

step 3.1L2L4algebra

5.1

Choose $z \in Q \cap (a - Q)$ . Write $z = [x^{2}]$ and $z = [a - y^{2}]$ for integers $x, y$ . Then $[x^{2} + y^{2}] = [a]$ by [L5], which is the required congruence.

step 4.1L5choosealgebradischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

39 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, Exercise 4 (standard reference, not scraped)