Linear Congruences

Definition 3.1 (Complete Set of Residues)   A complete set of residues modulo $n$ is a subset $R\subset\mathbb{Z}$ of size $n$ whose reductions modulo $n$ are distinct. In other words, a complete set of residues is a choice of representive for each equivalence class in $\mathbb{Z}/n\mathbb{Z}$.

Some examples:

$$R=\{0,1,2,\ldots,n-1\}$$

is a complete set of residues modulo $n$. When $n=5$, a complete set of residues is

$$R = \{0,1,-1,2,-2\}.$$

Lemma 3.2   If $R$ is a complete set of residues modulo $n$ and $a\in\mathbb{Z}$ with $\gcd(a,n)=1$, then $aR = \{ax : x \in R\}$ is also a complete set of residues.

Proof. If $ax\equiv ax'\pmod{n}$ with $x, x'\in R$, then Proposition 2.1 implies that $x\equiv {}x'\pmod{n}$. Because $R$ is a complete set of residues, this implies that $x=x'$. Thus the elements of $aR$ have distinct reductions modulo $n$. It follows, since $\char93 aR=n$, that $aR$ is a complete set of residues modulo $n$. $\qedsymbol$

Definition 3.3 (Linear Congruence)   A linear congruence is an equation of the form

$$ax\equiv b\pmod{n}.$$

Proposition 3.4   If $\gcd(a,n)=1$, then the equation

$$ax\equiv b\pmod{n}$$

must have a solution.

Proof. Let $R$ be a complete set of residues modulo $n$ (for example, $R=\{0,1,\ldots,n-1\}$). Then by Lemma 3.2, $aR$ is also a complete set of residues. Thus there is an element $ax\in aR$ such that $ax\equiv b\pmod{n}$, which proves the proposition. $\qedsymbol$

The point in the proof is that left multiplication by $a$ defines a map $\mathbb{Z}/n\mathbb{Z}\hookrightarrow \mathbb{Z}/n\mathbb{Z}$, which must be surjective because $\mathbb{Z}/n\mathbb{Z}$ is finite.

Illustration:

$$2x\equiv 3\pmod{7}$$

Set $R = \{0,1,2,3,4,5,6\}$. Then

$$2R = \{0,2,4,6,8\equiv 1, 10\equiv 3, 12\equiv 5\},$$

so $2\cdot 5\equiv 3\pmod{7}$.

Warning:
Note that the equation $ax\equiv b\pmod{n}$ might have a solution even if $\gcd(a,n)\neq 1$. To construct such examples, let $a$ be any divisor of $n$, $x$ any number, and set $b=ax$. For example, $2x\equiv 6\pmod{8}$ has a solution!