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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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For n1n\ge1, [a]n[a]_n is a unit if and only if gcd(a,n)=1\gcd(a,n)=1

Statement

Let n1n\ge1 and aZa\in\mathbb Z. Then [a]n[a]_n is a unit of Z/n\mathbb Z/n (The unit group (Z/n)×(\mathbb{Z}/n)^\times and Euler's totient φ(n)=(Z/n)×\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert for n1n\ge1) if and only if

gcd(a,n)=1,\gcd(a,n)=1,

that is, if and only if aa and nn are coprime (Coprime integers: gcd(a,b)=1\gcd(a,b) = 1). Consequently the condition gcd(a,n)=1\gcd(a,n)=1 depends only on the class [a]n[a]_n.

Facts & Assumptions

Given: A positive integer nn and an integer aa.

[F1]

The class [a]n[a]_n is a unit exactly when some [b]n[b]_n satisfies [a]n[b]n=[1]n[a]_n[b]_n=[1]_n (The unit group (Z/n)×(\mathbb{Z}/n)^\times and Euler's totient φ(n)=(Z/n)×\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert for n1n\ge1).

Proof

technique · direct
1.1

Suppose gcd(a,n)=1\gcd(a,n)=1. By [L1], choose integers x,yx,y with ax+ny=1ax+ny=1. Reducing this equality modulo nn gives [a]n[x]n=[1]n[a]_n[x]_n=[1]_n, so [a]n[a]_n is a unit.

L1F1choose
1.2

Conversely, suppose [a]n[a]_n is a unit. Choose bb with [a]n[b]n=[1]n[a]_n[b]_n=[1]_n. Then ab1(modn)ab\equiv1\pmod n, so ab1=nqab-1=nq for some integer qq, and ab+n(q)=1ab+n(-q)=1. By [L1], gcd(a,n)=1\gcd(a,n)=1.

F1L1choose
1.3

If aa(modn)a\equiv a'\pmod n and gcd(a,n)=1\gcd(a,n)=1, write a=a+nka'=a+nk. A Bézout identity ax+ny=1ax+ny=1 then becomes ax+n(ykx)=1a'x+n(y-kx)=1, so gcd(a,n)=1\gcd(a',n)=1 by [L1]; symmetry gives the converse. Thus the criterion is representative independent.

L1L2algebra
2.1

Steps 1.1 and 1.2 prove the equivalence, and step 1.3 proves its final representative-independence assertion.

step 1.1step 1.2step 1.3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 70 results over 18 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