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A unit is a square modulo exactly when it is a square at every prime-power factor
Statement
Let and let . A unit is a square modulo if and only if it is a square modulo every prime-power factor of .
Equivalently, for every odd prime one must have ; for the factor , there is no additional condition when , one needs when , and one needs when . At , the unique unit class is a square.
Facts & Assumptions
Given: A positive integer and a unit class .
For an odd prime , , and , the congruence is soluble if and only if (Unit square criterion and root count modulo odd prime powers).
For powers of , the unit square criterion is automatic modulo , is modulo , and is modulo for (Unit square criterion and root count modulo powers of two).
For pairwise coprime positive integers with product , the Chinese remainder map gives , including the empty list (For pairwise coprime positive moduli, the Chinese remainder bijection restricts to an isomorphism of unit groups).
The prime factors and exponents in the canonical factorisation of a positive integer are determined by that integer (For and any injective list of primes containing every prime divisor of , one has ; the exponents are determined by , and for every prime outside the list).
Every positive integer has a finite prime factorisation, unique up to the order of its prime factors (The fundamental theorem of arithmetic: every integer is a product of primes, and the factorisation is unique up to order — if with every and prime, then and for some ).
Proof
By [L5], has a finite prime factorisation; grouping equal factors and using [L4] gives its uniquely determined nontrivial prime-power factors, which are pairwise coprime. Apply [L3] to identify the unit group modulo with the product of their unit groups; when , this is the empty product of groups and has one element.
Under the isomorphism of step 1.1, a global square maps to a square in every component. Conversely, if every component is a square, choose one root in each of the finitely many nonempty local root sets and apply the inverse CRT isomorphism to obtain a global root. Substitution of [L1] and [L2] gives the explicit local conditions in the Statement, and the empty product handles .
Depends on
- Unit square criterion and root count modulo odd prime powers
- Unit square criterion and root count modulo powers of two
- For pairwise coprime positive moduli, the Chinese remainder bijection restricts to an isomorphism of unit groups
- For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n) = 0$ for every prime $q$ outside the list
- The fundamental theorem of arithmetic: every integer $n \ge 1$ 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 $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 114 results over 28 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)