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

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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.

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Unit square criterion and root count modulo powers of two

Statement

Let $a$ be odd.

Modulo $2$ , the congruence $x^{2} \equiv a (m o d 2)$ has exactly one solution class.
Modulo $4$ , it is soluble if and only if $a \equiv 1 (m o d 4)$ , and then it has exactly two solution classes.
For $k \geq 3$ , the congruence $x^{2} \equiv a (m o d 2^{k})$ is soluble if and only if $a \equiv 1 (m o d 8)$ ; when soluble, the number of roots is one for modulus $2$ , two for modulus $4$ , and four for modulus $2^{k}$ with $k \geq 3$ .

Facts & Assumptions

Given: An odd integer $a$ and an integer exponent $k \geq 1$ .

[L1]

For $k \geq 3$ , every unit modulo $2^{k}$ has a unique representation $(- 1)^{ε} 5^{j}$ with $ε \in {0, 1}$ and $j$ modulo $2^{k - 2}$ (For $k \geq 3$ , $(Z / 2^{k} Z)^{\times} ≅ C_{2} \times C_{2^{k - 2}}$ , generated uniquely as $(- 1)^{ε} 5^{j}$ ).

Proof

technique · direct

1.1givenalgebra

Modulo $2$ , the unique odd class is $1$ and its square is $1$ . Modulo $4$ , the odd classes $1$ and $3$ both square to $1$ , so an odd target is soluble exactly when it is $1$ modulo $4$ , and then both odd classes are roots.

1.2L1algebra

Let $k \geq 3$ and write a unit uniquely as $(- 1)^{ε} 5^{j}$ by [L1]. Squaring sends $(ε, j)$ to $(0, 2 j)$ , so a unit is a square exactly when $ε = 0$ and $j$ is even. Modulo $8$ , the four coordinate-parity possibilities give residues $1, 5, 7, 3$ , respectively, so this condition is equivalent to $a \equiv 1 (m o d 8)$ .

2.1step 1.1step 1.2L1algebra∎

The kernel of the squaring map in the coordinates of [L1] has the two choices for $ε$ and the two solutions of $2 j = 0$ modulo $2^{k - 2}$ , hence has four elements. Every nonempty fibre of a group homomorphism is a translate of its kernel, so every soluble target for $k \geq 3$ has exactly four roots. Together with step 1.1, this proves all criteria and counts.

Depends on

For $k\ge3$, $(\mathbb Z/2^k\mathbb Z)^\times\cong C_2\times C_{2^{k-2}}$, generated uniquely as $(-1)^\varepsilon5^j$

Used by

Dependency tree · two levels

7 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 (standard reference, not scraped)
V. Shoup, A Computational Introduction to Number Theory and Algebra, 2nd ed., §12.4 (standard reference, not scraped)