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The number of square roots of a unit modulo $n$ is the product of the local counts

Statement

Let $[a]_{n}$ be a unit square modulo $n \geq 1$ . The number of square roots is the product of the local root counts.

More explicitly, write $n = 2^{e} \prod_{i < r} p_{i}^{e_{i}}$ with distinct odd primes $p_{i}$ . The number of roots of $x^{2} \equiv a (m o d n)$ is $2^{r} c_{e}$ , where

$c_{e} = {\begin{cases} 1, & e = 0 or e = 1, \\ 2, & e = 2, \\ 4, & e \geq 3. \end{cases}$

For odd $n = \prod_{i < r} p_{i}^{e_{i}}$ , the soluble unit congruence $x^{2} \equiv a (m o d n)$ has exactly $2^{r}$ roots. At $n = 1$ the empty product gives one root.

Facts & Assumptions

Given: A positive integer $n$ and a unit $[a]_{n}$ for which $x^{2} \equiv a (m o d n)$ is soluble.

[L1]

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

[L2]

A soluble unit square congruence modulo $2^{e}$ has one root for $e = 1$ , two roots for $e = 2$ , and four roots for $e \geq 3$ (Unit square criterion and root count modulo powers of two).

[L3]

The Chinese remainder map is a group isomorphism from the unit group modulo a product of pairwise coprime positive integers to the product of their unit groups, including the empty product (For pairwise coprime positive moduli, the Chinese remainder bijection restricts to an isomorphism of unit groups).

[L4]

Every positive integer has a finite prime factorisation, unique up to the order of its prime factors (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ ).

Proof

technique · direct

1.1

By [L4], group the finite prime factorisation of $n$ into its pairwise coprime prime-power factors. Restrict the CRT isomorphism [L3] to the equation $u^{2} = [a]_{n}$ . It gives a bijection from the global root set to the Cartesian product of the root sets in those components, because an element satisfies the global equation exactly when each component satisfies its local equation.

L3L4given

2.1

The cardinality of that finite Cartesian product is the product of its local cardinalities. Each odd-prime-power factor contributes $2$ by [L1], while [L2] gives the factor $c_{e}$ for the two-part; if there are no prime-power factors, the empty product is $1$ . This is the displayed formula.

step 1.1L1L2algebra∎

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Sources

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

The number of square roots of a unit modulo nn is the product of the local counts