ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Lifting both square roots of $2$ from modulo $7$ to modulo $7^{4}$

Example

The two roots of $2$ lift along the powers of $7$ as follows:

modulus	first branch	second branch
$7$	$3$	$4$
$49$	$10$	$39$
$343$	$108$	$235$
$2401$	$2166$	$235$

Thus the square roots of $2$ modulo $7^{4} = 2401$ are the classes of $235$ and $2166$ .

Facts & Assumptions

Given: The target $a = 2$ , the odd prime $p = 7$ , and the two roots $3, 4$ modulo $7$ .

[L1]

A nonsingular root modulo $p^{k}$ has a unique lift of the form $x_{k} + t p^{k}$ modulo $p^{k + 1}$ , with $t$ uniquely determined modulo $p$ (A nonsingular square root lifts uniquely by one odd-prime-power step).

[L2]

A soluble unit square congruence modulo an odd prime power has exactly two solution classes (Unit square criterion and root count modulo odd prime powers).

Verification

technique · direct

1.1

Direct squaring gives the roots $3, 4$ modulo $7$ . Solving the correction congruence of [L1] on each branch gives corrections $(1, 5)$ modulo $7$ for the lift to $49$ , then $(2, 4)$ for the lift to $343$ , and finally $(6, 0)$ for the lift to $2401$ ; these yield the two displayed chains $3 \mapsto 10 \mapsto 108 \mapsto 2166$ and $4 \mapsto 39 \mapsto 235 \mapsto 235$ .

L1givenalgebra

2.1

One has $235^{2} = 23 \cdot 2401 + 2$ and $2166^{2} = 1954 \cdot 2401 + 2$ , so both final classes are roots. They are distinct, and [L2] says a soluble unit congruence has exactly two roots, so the list is complete.

step 1.1L2algebra∎

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Lifting both square roots of 22 from modulo 77 to modulo 7474