Exercise 1.8

Proof. Suppose $a\ne 0,b\ne 0$.

Let $x_0$ and $y_0$ be a solution to $\mathit{ax}+\mathit{by}=c$.

If $(x,y)$ is any solution of the same equation,

then

so

Let $a^{\prime }=a∕d,b^{\prime }=b∕d$ : from ex. 1.7, we know that $a^{\prime }\wedge b^{\prime }=1$.

As $a^{\prime }(x-x_0)=-b^{\prime }(y-y_0)$, $b^{\prime }\mid a^{\prime }(x-x_0)$, and $b^{\prime }\wedge a^{\prime }=1$, so (Gauss’ Lemma : prop. 1.1.1) $b^{\prime }\mid x-x_0$.

There exists $t\in \mathbb{Z}$ such that $x-x_0=t{b}^{\prime }$. Then $a^{\prime }t{b}^{\prime }=-b^{\prime }(y-y_0)$. As $b\ne 0$, $b^{\prime }\ne 0$, so $a^{\prime }t=-(y-y_0)$ :

Conversely, $a(x_0+t(b∕d))+b(y_0-t(a∕d))=a{x}_0+b{y}_0=c$.

Conclusion : if $a\ne 0,b\ne 0$, and $a{x}_0+b{y}_0=c$,

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