Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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 two reciprocity lattice counts partition an open rectangle

Statement

Let p and q be distinct odd primes, and let Sp,q and Sq,p be the lower-half lattice counts of Gauss's lemma as a lower-half lattice-point count. Then Sp,q+Sq,p=(p1)(q1)/4.

Facts & Assumptions

Given: Distinct odd primes p,q and the integer rectangle R={(x,y):1x(p1)/2, 1y(q1)/2}.

[L1]

Put Sp,q:={(x,y)Z2:1x(p1)/2, 0<py<qx} (Gauss's lemma as a lower-half lattice-point count).

[L2]

If a prime p divides a product ab, then pa or pb (Euclid's lemma: if p is prime and pab then pa or pb).

Proof

technique · direct
1.1

No point (x,y)R lies on the diagonal py=qx: equality would give pqx, so [L2] would give pq or px; distinctness of the primes rules out the first alternative, while 1x(p1)/2<p rules out the second. Thus every point of R satisfies exactly one of py<qx and qx<py.

L2given
2.1

The points of R with py<qx are exactly those counted by Sp,q: the inequality itself forces y<q/2, hence y(q1)/2; after interchanging the coordinates and the primes, the points with qx<py are exactly those counted by Sq,p. By step 1.1 these two sets partition R, whose cardinality is ((p1)/2)((q1)/2)=(p1)(q1)/4.

step 1.1L1algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 51 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources