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

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.

Unit square criterion and root count modulo odd prime powers

Statement

For an odd prime $p$ , $k \geq 1$ , and $p ∤ a$ , the congruence $x^{2} \equiv a (m o d p^{k})$ is soluble if and only if $(\frac{a}{p}) = 1$ .

When soluble it has exactly two solution classes modulo $p^{k}$ .

Facts & Assumptions

Given: An odd prime $p$ , an integer $k \geq 1$ , and an integer $a$ with $p ∤ a$ .

[L1]

Every root modulo $p^{j}$ has a unique lift to a root modulo $p^{j + 1}$ when $j \geq 1$ and $p ∤ a$ (A nonsingular square root lifts uniquely by one odd-prime-power step).

[L2]

The congruence $x^{2} \equiv a (m o d p)$ has exactly $1 + (a / p)$ solution classes modulo $p$ ( $x^{2} \equiv a (m o d p)$ has exactly $1 + (a / p)$ solution classes).

[L3]

For $p ∤ a$ , the Legendre symbol $(a / p)$ is $1$ when $a$ is a square modulo $p$ and $- 1$ otherwise (The Legendre symbol, including its zero value).

Proof

technique · direct

1.1L2L3given

Any root modulo $p^{k}$ reduces to a root modulo $p$ . Since $p ∤ a$ , [L3] makes $(a / p)$ a sign, and [L2] says that a root exists only when $1 + (a / p) = 2$ , equivalently when $(a / p) = 1$ .

1.2L1L2L3

Conversely, if $(a / p) = 1$ , [L2] gives exactly two root classes modulo $p$ . For $k = 1$ these are the required roots. For $k > 1$ , repeatedly apply [L1] from exponent $1$ through exponent $k - 1$ to lift each class uniquely; the two lifted classes remain distinct because their reductions modulo $p$ are distinct.

2.1step 1.1step 1.2L1∎

Every root modulo $p^{k}$ reduces to one of the two roots modulo $p$ , and at every successive exponent [L1] forces it to be the unique lift of that reduction. Thus step 1.2 constructs all roots, so there are exactly two. Together with step 1.1 this proves both directions of the criterion and the count.

Depends on

Used by

Dependency tree · two levels

12 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, §§B.VII and D.I (standard reference, not scraped)
A. Gorodnik, Number Theory, Lecture 7, §1 (standard reference, not scraped)
V. Shoup, A Computational Introduction to Number Theory and Algebra, 2nd ed., §12.4 (standard reference, not scraped)