Given the equation, $y = \frac{3(x-5)}{(x-7)}$, how do I find the set of solutions $(x, y)$ given $x, y \in \mathbb{Z^+}$, and prove that no other solutions exist.
I am able to do the first part by simply choosing arbitrary $x$ values. For example, integer solutions include the following: $(1, 2), (4, 1), (8, 9), (9, 6), $ etc.
How do I formally prove that these are solutions, and that no other solutions in the positive integers exist for this equation?
We have: $y = 3+ \dfrac{6}{x-7}\implies x-7 \mid 6\implies x-7 = 1,-1,2,-2,3,-3,6,-6\implies x = ...., y = ....$