TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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Square criterion in Q_2

Statement

Let $x \in Q_{2}^{\times}$ . Write

$x = 2^{n} u$

with $n \in Z$ and odd $u \in Z_{2}^{\times}$ . Then $x$ is a square in $Q_{2}$ if and only if $n$ is even and $u \equiv 1 (m o d 8)$ .

Facts & Assumptions

Given: $x = 2^{n} u$ with odd $u \in Z_{2}^{\times}$ .

[L1]

$Z_{2}$ is the valuation ring of $Q_{2}$ (Z_p is the valuation ring of Q_p).

[L2]

Newton's criterion holds in $Q_{2}$ (Newton's criterion in Q_p).

Proof

technique · direct

1.1L1givenalgebra

If $x = y^{2}$ , write $y = 2^{m} v$ with odd $v \in Z_{2}^{\times}$ . Then $n = 2 m$ is even. Every odd square is congruent to $1$ modulo $8$ , so the unit factor $u$ is congruent to $1$ modulo $8$ .

1.2L2givenalgebra

Conversely, assume $n = 2 m$ and $u \equiv 1 (m o d 8)$ . For $f (X) = X^{2} - u$ at $a_{0} = 1$ , $∣ f (1) ∣_{2} = ∣ 1 - u ∣_{2} \leq 2^{- 3} < 2^{- 2} = ∣ f^{'} (1) ∣_{2}^{2} .$ By [L2], Newton iteration converges to a root $v \in Z_{2}$ of $f$ , so $v^{2} = u$ . Then $y : = 2^{m} v$ satisfies $y^{2} = x$ .

2.1step 1.1step 1.2∎

This proves both directions of the criterion.

Depends on

Used by

The two-adic square test separates 17 and 5 Example
There is no square root of p in Qₚ Example
Hasse-Minkowski theorem over Q Theorem
The two-adic Hilbert symbol formula 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

Keith Conrad, Hensel's Lemma, Theorem 4.5 (standard reference, not scraped)
Andrew V. Sutherland, 18.782 Lecture 10 (standard reference, not scraped)