LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-17

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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.

Gauss's lemma as a lower-half lattice-point count

Statement

Let $p$ and $q$ be distinct odd primes. Put $S_{p, q} : = ∣ {(x, y) \in Z^{2} : 1 \leq x \leq (p - 1) / 2, 0 < p y < q x} ∣$ .

Then $(\frac{q}{p}) = (- 1)^{S_{p, q}}$ .

Thus Gauss's sign count (Gauss's quadratic-residue lemma) can be read as the parity of a finite set of lattice points, without introducing floor notation.

Facts & Assumptions

Given: Distinct odd primes $p, q$ , and $m = (p - 1) / 2$ .

[L1]

If $N (q, p)$ is the number of least positive residues of $q, 2 q, \dots, m q$ modulo $p$ that exceed $p / 2$ , then $(\frac{q}{p}) = (- 1)^{N (q, p)}$ (Gauss's quadratic-residue lemma).

[L2]

For every integer $A$ and positive integer $p$ , there are unique integers $t, ρ$ such that $A = p t + ρ$ and $0 \leq ρ < p$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L3]

For each $1 \leq x \leq m$ , there are unique $ε_{x} \in {1, - 1}$ and $u_{x} \in {1, \dots, m}$ such that $q x \equiv ε_{x} u_{x} (m o d p)$ , and $u_{1}, \dots, u_{m}$ is a permutation of $1, \dots, m$ (Multiplication by $a$ with $p ∤ a$ permutes an odd prime's signed half-system up to sign).

Proof

technique · direct

1.1L2givenalgebra

For each $1 \leq x \leq m$ , [L2] gives $q x = p t_{x} + ρ_{x}$ with $0 \leq ρ_{x} < p$ ; since $p ∤ q x$ , one has $0 < ρ_{x} < p$ . The positive integers $y$ satisfying $p y < q x = p t_{x} + ρ_{x}$ are exactly $1, \dots, t_{x}$ , so $S_{p, q} = \sum_{x = 1}^{m} t_{x}$ .

2.1step 1.1L1L3algebra∎

In the notation of [L3], $ρ_{x} = u_{x}$ when $ε_{x} = 1$ and $ρ_{x} = p - u_{x}$ when $ε_{x} = - 1$ ; the negative signs are exactly the residues counted by $N (q, p)$ . Summing the equations of step 1.1 and reducing modulo $2$ gives $\sum_{x} x \equiv S_{p, q} + \sum_{x} ρ_{x} (m o d 2)$ , because $p$ and $q$ are odd. Also $\sum_{x} ρ_{x} \equiv \sum_{x} u_{x} + N (q, p) (m o d 2)$ , while the permutation in [L3] gives $\sum_{x} u_{x} = \sum_{x} x$ . Hence $S_{p, q} \equiv N (q, p) (m o d 2)$ , and [L1] yields $(\frac{q}{p}) = (- 1)^{S_{p, q}}$ .

Depends on

Used by

Dependency tree · two levels

24 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.V (standard reference, not scraped)
A. Gorodnik, Number Theory, Lecture 9, §2 (standard reference, not scraped)