LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06

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.

Primitive Gauss-sum twist

Statement

If $χ$ is primitive modulo $q$ , then for every integer $m$ , $\sum_{a m o d q} χ (a) e (a m / q) = \overline{χ (m)} τ (χ) .$

Facts & Assumptions

Given: A primitive $χ$ modulo $q$ and $m \in Z$ .

[F1]

The normalization of $τ (χ)$ is fixed above (Gauss sum of a Dirichlet character).

[F2]

Primitivity excludes induction from a proper divisor (Primitive Dirichlet characters and conductor).

Proof

technique · cases

1.1F1givenassume-case unit

If $(m, q) = 1$ , substitution $b = a m$ permutes the residue classes and gives the sum as $\overline{χ (m)} τ (χ)$ .

2.1F2step 1.1assume-case nonunit

If $g = (m, q) > 1$ , choose a unit $u \equiv 1 (m o d q / g)$ for which $χ (u) \neq 1$ ; otherwise the unit character factors modulo $q / g$ , contrary to [F2]. Multiplication by $u$ leaves $e (a m / q)$ unchanged and multiplies the sum by $χ (u)$ , so the sum is zero.

3.1step 2.1cases-exhaustive∎

In the second case $χ (m) = 0$ , so the right side is also zero; the two cases prove the formula.

Depends on

Used by

Norm of a primitive Gauss sum Theorem
Twisted Poisson summation Theorem

Dependency tree · two levels

4 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

Nickolas Andersen, Analytic Number Theory, Lemma 16.3 (standard reference, not scraped)