The question is:
Find all pairs of integers $(x, y)$ such that $61x+18y = 0$.
By solving for $x$, we have $x = (-18/61) \cdot y$, therefore $x$ is an integer if-f $y$ is a multiple of $61$.
Similarly, by solving for $y$, we have $y = (-61/18) \cdot x$, therefore $y$ is an integer if-f $x$ is a multiple of $18$.
Therefore, the solutions are of the form $(x,y)$, where $x = 18k$ and $y =(-61/18)\cdot x$, for all integer $k$.
Is this right? Am I missing something? Thanks in advance!
I think you can do better. Your characterization is problematic because you cannot know in advance, before arithmetic, which $k$ will fit.
Consider that $\gcd(18,61) = 1$, which means that starting with $k=0 \iff x=y=0$, you must add multiples of $18 \cdot 61$ to make things work. The idea would be to write a characterization to work for every $k$...