TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Quadratic reciprocity for coprime odd Jacobi denominators

Statement

For coprime odd positive integers $m, n$ ,

$(\frac{m}{n}) (\frac{n}{m}) = (- 1)^{(m - 1) (n - 1) / 4} .$

The formula includes $m = 1$ or $n = 1$ .

Facts & Assumptions

Given: Coprime odd positive integers $m, n$ .

[L1]

For distinct odd primes $p, q$ , $(\frac{p}{q}) (\frac{q}{p}) = (- 1)^{(p - 1) (q - 1) / 4}$ (Quadratic reciprocity for distinct odd primes).

[L2]

For an odd positive denominator, the Jacobi symbol is the product of the Legendre symbols over its canonical prime factors with multiplicity (The Jacobi symbol, with its zero value and empty-product convention).

[L3]

The Jacobi symbol is multiplicative in both its numerator and its odd positive denominator (The Jacobi symbol is multiplicative in numerator and denominator).

[L4]

The primes and their exponents in the canonical factorisation of a positive integer are determined by that integer (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).

[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

1.1L1L2L3L4L5givenalgebra

By [L5], $m$ and $n$ have finite prime factorisations; grouping equal factors and using [L4], write their canonical forms as $m = \prod_{i} p_{i}^{α_{i}}$ and $n = \prod_{j} q_{j}^{β_{j}}$ . Expanding both Jacobi symbols by [L2] and [L3] expresses their product as $\prod_{i, j} {[(\frac{p_{i}}{q_{j}}) (\frac{q_{j}}{p_{i}})]}^{α_{i} β_{j}}$ . Coprimality makes every $p_{i}$ distinct from every $q_{j}$ , so [L1] turns this into $(- 1)^{E}$ , where $E = \sum_{i, j} α_{i} β_{j} (p_{i} - 1) (q_{j} - 1) / 4$ .

2.1step 1.1L2algebra∎

For a product of odd integers, repeated use of $(u v - 1) / 2 \equiv (u - 1) / 2 + (v - 1) / 2 (m o d 2)$ gives $(m - 1) / 2 \equiv \sum_{i} α_{i} (p_{i} - 1) / 2$ and $(n - 1) / 2 \equiv \sum_{j} β_{j} (q_{j} - 1) / 2$ modulo $2$ . Their product is congruent to $E$ , so step 1.1 gives the stated sign. If either integer is $1$ , the relevant prime list and cross-product are empty and both sides equal $1$ .

Depends on

Used by

The Euclidean algorithm computes the Jacobi symbol without factoring the denominator Theorem

Dependency tree · two levels

44 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, §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)