Proposition 3

Let $a, b, c$ be elements of the field of fractions of a principal ideal domain $A$, and let $d$ be a gcd of $a$ and $c$; then the congruence $ax \equiv b \pmod{c}$ has a solution $x_0 \in A$ if and only if $d$ divides $b$; in this case the elements $x$ of $A$ which satisfy $ax \equiv b \pmod{c}$ are precisely those which satisfy $x \equiv x_0 \pmod{cd}$.

If $ax \equiv b \pmod{c}$, with $x \in A$, then there exists $y \in A$ such that $b = ax + cy$, so $d$ divides $b$. Conversely, if $d$ divides $b$ then $b = ax_0 + cy_0$ with $x_0, y_0 \in A$ (Prop. 1), so $ax_0 \equiv b \pmod{c}$; moreover, the relation $ax \equiv b \pmod{c}$ is then equivalent to $a(x - x_0) \equiv 0 \pmod{c}$; putting $a = da'$ and $c = dc'$, the latter becomes $a'(x - x_0) \equiv 0 \pmod{c'}$. But this is equivalent (for $x \in A$) to $x - x_0 \equiv 0 \pmod{c'}$, since $a'$ and $c'$ are coprime (VI, p. 14, Prop. 10 (DIV) and VI, p. 15, Cor. 2 to Prop. 11 (DIV)).