LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.

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Multiplication by $a$ with $p ∤ a$ permutes an odd prime's signed half-system up to sign

Statement

Let $p$ be an odd prime, let $p ∤ a$ , and put $m = (p - 1) / 2$ . For each $1 \leq j \leq m$ , there are unique $ε_{j} \in {1, - 1}$ and $r_{j} \in {1, \dots, m}$ such that

$a j \equiv ε_{j} r_{j} (m o d p) .$

The absolute representatives $r_{1}, \dots, r_{m}$ are a permutation of $1, \dots, m$ .

Facts & Assumptions

Given: An odd prime $p$ , an integer $a$ with $p ∤ a$ , and $m = (p - 1) / 2$ .

[L1]

Every class modulo a positive integer $n$ has exactly one representative $s$ with $0 \leq s < n$ (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L2]

The quotient $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L3]

Every injection from a finite set to itself is a bijection (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

Proof

technique · direct

1.1L1L2given

For $1 \leq j \leq m$ , [L1] gives the standard representative $s_{j}$ of $a j$ . It is nonzero because [L2] permits cancellation of the nonzero classes $[a]_{p}$ and $[j]_{p}$ . If $s_{j} \leq m$ , set $(ε_{j}, r_{j}) = (1, s_{j})$ ; if $s_{j} > m$ , set $(ε_{j}, r_{j}) = (- 1, p - s_{j})$ . Since $p = 2 m + 1$ , this gives the stated unique signed representative with $1 \leq r_{j} \leq m$ .

2.1L2step 1.1

If $r_{j} = r_{k}$ , then $a j \equiv a k$ or $a j \equiv - a k (m o d p)$ . Cancelling $[a]_{p}$ by [L2] gives $j \equiv k$ or $j \equiv - k (m o d p)$ . In the first case $∣ j - k ∣ < p$ forces $j = k$ ; in the second, $2 \leq j + k \leq p - 1$ , so $p ∤ (j + k)$ , a contradiction. Thus $j \mapsto r_{j}$ is injective.

3.1L3step 2.1∎

The map $j \mapsto r_{j}$ is an injection from the finite set ${1, \dots, m}$ to itself, so [L3] makes it a bijection. Therefore $r_{1}, \dots, r_{m}$ is a permutation of the half-system.

Depends on

Used by

The signed half-system for 3 modulo 11 gives (3/11)=1 Example
Gauss's lemma as a lower-half lattice-point count Lemma
Gauss's quadratic-residue lemma Theorem

Dependency tree · two levels

25 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

H. Hackman, Elementary Number Theory, Chapter D, Section D.IV (standard reference, not scraped)
W. Stein, Elementary Number Theory, Section 4.3 (standard reference, not scraped)

Multiplication by aa with p∤ap∤a permutes an odd prime's signed half-system up to sign