if $k>1$, Does $a+b =k(ax+by)$ have finitely many solutions?

Question

Let $a,b,k,x,y$ be non-zero integers, solve $a+b=k(ax+by)$. It's a rather simple problem, but I just want to make sure that I have got all the possible solutions.

Answer 1

I'm assuming you are given fixed numbers $a$ and $b$, let's further assume that $a+b\ne0$ (otherwise it's an easy problem). Since $k\gcd(a,b)$ divides the right-hand side, a necessary condition for solutions to exist is that $k\gcd(a,b)\mid (a+b)$; equivalently, that $k$ divides $(a+b)/\gcd(a,b)$.

So now fix one of the finitely many integers $k$ dividing $(a+b)/\gcd(a,b)$. By Bezout's identity, there is a solution $(x_0,y_0)$ to $ax+by = \gcd(a,b)$; with that solution in hand, it's easy to construct infinitely many solutions to $ax+by = \gcd(a,b)$, namely $(x,y) = (x_0,y_0) + (tb/\gcd(a,b), -ta/\gcd(a,b))$. Now multiply any of these solutions by $(a+b)/k\gcd(a,b)$ to obtain solutions to $ax+by = (a+b)/k$.