Infinitude of solutions to Bézout's equation If $c=\gcd(a,b)$ for $a,b\in\Bbb Z$ it is possible to write $ax + by = c$ with $x,y\in\Bbb Z$. Is it possible to prove that there are infinitely many integer solutions for $x$ and $y$ without using Diophantine equations, and if so how?
(The particular problem I have is $1947x+264y=33$.)
 A: Writing $ax + by = c$, you get infinitely many solutions as follows: Let $k\in \mathbb Z$ be arbitrary, then
\begin{align*}
c &= ax + by = ax + \frac{ab}{c}\cdot k + by - \frac{ab}{c}\cdot k\\
&= a\cdot \left(x+\frac{b}{c}\cdot k\right) + b\cdot \left(y-\frac{a}{c}\cdot k\right) \tag{1}
\end{align*}
is another solution. Conversely, every solution is of this form: Take two solutions $ax + by = c$ and $au + bv = c$. Subtracting both equations, dividing by $c$ and rearranging yields
$$
\frac{a}{c}\cdot (x-u) = \frac{b}{c}\cdot (v-y). \tag{2}
$$
Since $\frac{a}{c}$ and $\frac{b}{c}$ are coprime, it follows that $\frac{b}{c}$ divides $x-u$ and, likewise, $\frac{a}{c}$ divides $v-y$. Hence, we find $n,m\in \mathbb Z$ with $\frac{a}{c}\cdot n = v-y$ and $\frac{b}{c}\cdot m = x-u$. Plugging both solutions into (2) gives
$$
\frac{a}{c}\cdot \frac{b}{c}\cdot m = \frac{b}{c}\cdot\frac{a}{c}\cdot n
$$
and hence $n=m$, because $\frac{ab}{c^2}\neq 0$. This shows that every solution is of the form (1).
