LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-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.

The Brahmagupta–Fibonacci two-square identity

Statement

For all integers $a, b, c, d$ ,

$(a^{2} + b^{2}) (c^{2} + d^{2}) = (a c - b d)^{2} + (a d + b c)^{2} = (a c + b d)^{2} + (a d - b c)^{2} .$

Facts & Assumptions

Given: Integers $a, b, c, d$ .

Proof

technique · direct

1.1

Expanding gives $(a c - b d)^{2} + (a d + b c)^{2} = a^{2} c^{2} - 2 a b c d + b^{2} d^{2} + a^{2} d^{2} + 2 a b c d + b^{2} c^{2} = (a^{2} + b^{2}) (c^{2} + d^{2})$ .

givenalgebra

2.1

Likewise $(a c + b d)^{2} + (a d - b c)^{2} = a^{2} c^{2} + 2 a b c d + b^{2} d^{2} + a^{2} d^{2} - 2 a b c d + b^{2} c^{2} = (a^{2} + b^{2}) (c^{2} + d^{2})$ .

givenalgebra∎

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

P. Hackman, Elementary Number Theory, Chapter E, §E.II.1 (standard reference, not scraped)
W. Stein, Elementary Number Theory: Primes, Congruences, and Secrets, §5.7 (standard reference, not scraped)