Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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 supplementary laws for the Jacobi symbol

Statement

For every odd positive integer n,

(−1n)=(−1)(n−1)/2,(2n)=(−1)(n2−1)/8.

Both formulas include n=1, where each Jacobi symbol and each displayed power of −1 equals 1.

Facts & Assumptions

Given: An odd positive integer n.

[L1]

The Jacobi symbol is the product of the prime Legendre symbols, taken with the multiplicities in the canonical prime factorisation (The Jacobi symbol, with its zero value and empty-product convention).

[L2]

For odd positive u,v, (auv)=(au)(av) (The Jacobi symbol is multiplicative in numerator and denominator).

[L3]

For every odd prime p, (−1p)=(−1)(p−1)/2 (First supplement: (−1/p)=(−1)(p−1)/2).

[L4]

For every odd prime p, (2p)=(−1)(p2−1)/8 (Second supplement: (2/p)=(−1)(p2−1)/8).

Proof

technique · direct
1.1L1L2L3algebra

Expand (−1n) through [L1], applying [L3] to every prime factor with multiplicity and [L2] to multiply the contributions. For odd u,v, the difference (uv−1)/2−(u−1)/2−(v−1)/2=(u−1)(v−1)/2 is even, so iteration through the factor list gives (−1n)=(−1)(n−1)/2; for the empty factor list n=1, both sides are 1.

2.1L1L2L4algebra∎

Similarly, [L1] and [L4] give the product of the signs (−1)(p2−1)/8 with multiplicity. For odd u,v, the difference ((uv)2−1)/8−(u2−1)/8−(v2−1)/8=(u2−1)(v2−1)/8 is even, because each of u2−1 and v2−1 is divisible by 8. Iterating this identity gives (2n)=(−1)(n2−1)/8, again with value 1 at n=1.

Depends on

Used by

Dependency tree · two levels

15 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