TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

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.

Square criterion in Q_p for odd p

Statement

Let $p$ be odd and let $x \in Q_{p}^{\times}$ . Write

$x = p^{n} u$

with $n \in Z$ and $u \in Z_{p}^{\times}$ . Then $x$ is a square in $Q_{p}$ if and only if $n$ is even and the reduction of $u$ in $F_{p}^{\times}$ is a square.

Facts & Assumptions

Given: An odd prime $p$ and $x = p^{n} u$ with $u \in Z_{p}^{\times}$ .

[L1]

$Z_{p}$ is the valuation ring of $Q_{p}$ (Z_p is the valuation ring of Q_p).

[L2]

Simple roots lift uniquely in $Z_{p}$ (Simple roots lift uniquely in Z_p).

Proof

technique · direct

1.1

If $x = y^{2}$ , write $y = p^{m} v$ with $v \in Z_{p}^{\times}$ . Then $x = p^{2 m} v^{2}$ , so $n = 2 m$ is even and the reduction of $u$ is the square of the reduction of $v$ in $F_{p}^{\times}$ .

L1givenalgebra

1.2

Conversely, assume $n = 2 m$ and the residue class of $u$ is $c^{2}$ in $F_{p}^{\times}$ . Choose $a_{0} \in Z_{p}$ with $a_{0} \equiv c (m o d p)$ . For $f (X) = X^{2} - u$ , one has $f (a_{0}) \equiv 0 (m o d p)$ and $f^{'} (a_{0}) = 2 a_{0} \equiv̸ 0 (m o d p)$ because $p$ is odd and $c \neq 0$ . By [L2], $f$ has a root $v \in Z_{p}$ with $v^{2} = u$ . Then $y : = p^{m} v$ satisfies $y^{2} = x$ .

L2givenalgebra

2.1

Step 1.1 proves necessity and step 1.2 proves sufficiency.

step 1.1step 1.2∎

Depends on

Used by

A square root of minus one in Q₅ Example
There is no square root of p in Qₚ Example

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

Keith Conrad, Hensel's Lemma, Examples 4.3 and 4.4 (standard reference, not scraped)
Andrew V. Sutherland, 18.782 Lecture 10 (standard reference, not scraped)