ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-16

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.

A nonmonic quadratic congruence solved through its discriminant

Example

The congruence

$3 x^{2} + 4 x + 1 \equiv 0 (m o d 11)$

has exactly the two solution classes $[7]_{11}$ and $[10]_{11}$ .

Facts & Assumptions

Given: The polynomial $3 x^{2} + 4 x + 1$ modulo the odd prime $11$ .

[L1]

If $p$ is odd and $p ∤ A$ , then $A x^{2} + B x + C \equiv 0 (m o d p)$ has exactly $1 + ((B^{2} - 4 A C) / p)$ solution classes (The discriminant counts roots of $A x^{2} + B x + C \equiv 0 (m o d p)$ for odd prime $p ∤ A$ ).

[L2]

For an odd prime $p$ , $(\frac{a}{p}) = 1$ when $p ∤ a$ and $a$ is a quadratic residue modulo $p$ , and $(\frac{a}{p}) = - 1$ when $p ∤ a$ and $a$ is a quadratic nonresidue modulo $p$ (The Legendre symbol, including its zero value).

[L3]

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

Verification

technique · direct

1.1L2givenalgebra

The discriminant is $Δ = 4^{2} - 4 \cdot 3 \cdot 1 = 4$ . Here $11 ∤ 4$ and $2^{2} = 4$ , so $4$ is a quadratic residue modulo $11$ and [L2] gives $(Δ / 11) = 1$ .

2.1L1step 1.1

Since $11$ is odd and $11 ∤ 3$ , fact [L1] applies and predicts exactly $1 + 1 = 2$ solution classes.

3.1L3step 1.1step 2.1algebra∎

Completing the square gives $(6 x + 4)^{2} \equiv 4 (m o d 11)$ , so $((6 x + 4) - 2) ((6 x + 4) + 2) \equiv 0 (m o d 11)$ ; by [L3] the field $Z / 11$ has no zero divisors, so $6 x + 4 \equiv 2$ or $6 x + 4 \equiv - 2$ . Since $6^{- 1} \equiv 2 (m o d 11)$ , these yield $x \equiv 7$ and $x \equiv 10$ ; direct substitution gives $176 \equiv 0$ and $341 \equiv 0 (m o d 11)$ . Thus both predicted classes occur.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

A. Gorodnik, Number Theory, Lecture 9, Section 1 (standard reference, not scraped)