LemmaStatement: AI-adaptedProof: 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.

A nonsingular square root lifts uniquely by one odd-prime-power step

Statement

Let $p$ be an odd prime, let $k \geq 1$ , and let $a, x_{k} \in Z$ satisfy $p ∤ a$ and $x_{k}^{2} \equiv a (m o d p^{k})$ . Then there is a unique class $t (m o d p)$ such that $(x_{k} + t p^{k})^{2} \equiv a (m o d p^{k + 1})$ .

Equivalently, the root class of $x_{k}$ modulo $p^{k}$ has exactly one lift to a root class modulo $p^{k + 1}$ .

Facts & Assumptions

Given: An odd prime $p$ , an integer $k \geq 1$ , and integers $a, x_{k}$ with $p ∤ a$ and $x_{k}^{2} \equiv a (m o d p^{k})$ .

[L1]

If $d = \gcd (c, n)$ , then $c t \equiv b (m o d n)$ is soluble exactly when $d ∣ b$ , and when soluble it has exactly $d$ solution classes modulo $n$ (For $n \geq 1$ , $a x \equiv b (m o d n)$ is solvable exactly when $\gcd (a, n) ∣ b$ , and then has exactly $\gcd (a, n)$ solution classes modulo $n$ ).

[L2]

If a prime $p$ divides a product $u v$ , then $p ∣ u$ or $p ∣ v$ (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ ).

[L3]

If a prime $p$ does not divide an integer $u$ , then $\gcd (p, u) = 1$ (For a prime $p$ and any integer $a$ , $\gcd (p, a)$ is $p$ when $p ∣ a$ and $1$ otherwise; so $p ∤ a$ makes $p$ and $a$ coprime).

Proof

technique · direct

1.1L2L3given

Reducing the given congruence modulo $p$ shows that $p ∣ x_{k}$ would force $p ∣ a$ , contrary to the hypothesis. If $p ∣ 2 x_{k}$ , [L2] would give $p ∣ 2$ or $p ∣ x_{k}$ ; both are impossible because $p$ is odd. Hence $p ∤ 2 x_{k}$ , and [L3] gives $\gcd (2 x_{k}, p) = 1$ .

2.1step 1.1L1algebra∎

Write $x_{k}^{2} - a = c p^{k}$ . Every class modulo $p^{k + 1}$ reducing to $x_{k}$ modulo $p^{k}$ has a unique form $x_{k} + t p^{k}$ with $t$ modulo $p$ , and expansion gives $(x_{k} + t p^{k})^{2} - a = p^{k} (c + 2 x_{k} t + t^{2} p^{k})$ . Since $k \geq 1$ , the lift is a root modulo $p^{k + 1}$ exactly when $2 x_{k} t \equiv - c (m o d p)$ . By step 1.1 and [L1], this linear congruence has exactly one solution class $t$ modulo $p$ .

Depends on

Used by

Dependency tree · two levels

24 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 7, §1 (standard reference, not scraped)
P. Hackman, Elementary Number Theory, §B.VII (standard reference, not scraped)