Let $a ≠ 0$ and consider the linear equation $ax + may= na$. Prove that if $x_0, y_0$ is a particular solution of this equation then the general solution is given by $x = x_0 + mt$ and $y= y_0− t$.
By Proposition (1.18), the general solution is
$$ x = x_0 + \frac{ma}{\gcd(a,ma)}t \quad y = y_0 - \frac{a}{\gcd(a,ma)}t $$
for any integer $t$.
Let's consider $\gcd(a,ma)$
$$\begin{align} \gcd(a,ma) & = a \gcd(1,m) \quad \text{Proposition 1.11} \\ \\ & = a \end{align}$$
So the general solution is
$$ x = x_0 + mt \quad y = y_0 - t $$
for any integer $t$.