How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$? How can you prove that $b=2, m=3$ are the only positive integer solutions to $4b+3m=17$ without a proof by exhaustion?
$$ 4b+3m=17 \implies 4b=17-3m $$

$$\implies 4b=16+(m+1)-4m = 4(4-m)+ (m+1) $$

Thus $m+1$ must be a multiple of $4$ where $m&lt;4$ to make  both sides positive.

The only choice is $m=3$ which implies $b=2$
If $b\geq 9$ then (since $m\geq 1$) $$4b+3m&gt;18+3m&gt;17$$
If $m\geq 6$ then (since $b\geq 1$) $$4b+3m&gt;4b+18&gt;17$$


Thus there are only finitely many solutions, $1\leq b&lt; 9$, $1\leq m&lt; 6$, and they can be checked directly.