How to prove if $px+qy=r$ has an integer solution, then it has infinite integer solutions $(x,y)$ with $x$ being even or $y$ being even

Question

I only managed to come up with

$ A \Rightarrow B \lor C \equiv A \land\lnot B \Rightarrow C$

However, the negation of B is "there are finite integer solutions (x,y) where x is even" which doesn't help at all.

Need a hint

Answer 1

Note that $0 = pq-qp = -pq+qp$ so if $(x_0,y_0)$ is a solution, so are $(x_0+q, y_0-p)$ and $(x-q,y+p)$

Also let $2^s$ be the largest power of $2$ that divides both $p$ and $q$. Then set $a=p/2^s$ and $b=q/2^s$, and we know that at least one of $a$ and $b$ is odd and that $pb-qa=0$

Now $(x_0+kb, y_0-ka)$ are solutions for all $k \in \mathbb Z$, and at least one of these terms is changing parity as $k$ changes. So there are an infinite number of solutions with at least one even term, as required.