TheoremStatement: 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 Jacobi symbol is multiplicative in numerator and denominator

Statement

For integers $a, b$ and an odd positive integer $n$ ,

$(\frac{a b}{n}) = (\frac{a}{n}) (\frac{b}{n}) .$

For an integer $a$ and odd positive integers $m, n$ ,

$(\frac{a}{m n}) = (\frac{a}{m}) (\frac{a}{n}) .$

No coprimality hypothesis is imposed on either pair of arguments.

Facts & Assumptions

Given: Integers $a, b$ and odd positive integers $m, n$ .

[L1]

For odd $n \geq 1$ with canonical prime factorisation $n = \prod_{i < r} p_{i}^{e_{i}}$ , the Jacobi symbol is $(\frac{a}{n}) = \prod_{i < r} {(\frac{a}{p_{i}})}^{e_{i}}$ (The Jacobi symbol, with its zero value and empty-product convention).

[L2]

For every odd prime $p$ and integers $a, b$ , $(\frac{a b}{p}) = (\frac{a}{p}) (\frac{b}{p})$ (The Legendre symbol is multiplicative for all integer numerators).

[L3]

Canonical prime-factor exponents are determined by the positive integer being factored (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

[L4]

For a prime $p$ and nonzero integers $u, v$ , $v_{p} (u v) = v_{p} (u) + v_{p} (v)$ ( $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero).

[L5]

Every positive integer has a finite prime factorisation, unique up to the order of its prime factors (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ ).

Proof

technique · direct

1.1

Apply [L2] to every prime factor in [L1] and regroup the finite product: $(\frac{a b}{n}) = \prod_{i} {(\frac{a}{p_{i}})}^{e_{i}} \prod_{i} {(\frac{b}{p_{i}})}^{e_{i}} = (\frac{a}{n}) (\frac{b}{n})$ . This remains valid when a Legendre factor is zero.

L1L2algebra

2.1

By [L5], choose finite prime factorisations of $m$ and $n$ ; concatenating their factor lists gives a prime factorisation of $m n$ . Grouping equal primes and using [L3] and [L4], the exponent of each prime in $m n$ is the sum of its exponents in $m$ and $n$ . Substituting those sums in [L1] and regrouping gives $(\frac{a}{m n}) = (\frac{a}{m}) (\frac{a}{n})$ . If either denominator is $1$ , its factor list is empty and its contribution is $1$ .

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Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 97 results over 27 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

P. Hackman, Elementary Number Theory, §D.II (standard reference, not scraped)
A. Gorodnik, Number Theory, Lecture 10, §1 (standard reference, not scraped)
V. Shoup, A Computational Introduction to Number Theory and Algebra, 2nd ed., §12.2 (standard reference, not scraped)