No positive integers $a,b$ exist such that $7ab = 17(a+b)$

Question

I'm trying to prove the following:

No positive integers $a,b$ exist such that $7ab = 17(a+b)$

I've been struggling on this problem for the last hour and a half, and I keep running in circles. I've tried four or five different approaches, mostly revolving around divisibility / fundamental theorem of arithmetic, but I can't quite get the proof. I know that if this relationship held, then $ab = 17k$ and $a+b = 7k$ for some $k \in \mathbb{Z}^+$, but I can't seem to derive a contradiction from this. Can someone please help me finish this thing?

Answer 1

Without loss of generality, assume $a=17k$, then: $$7\cdot 17k\cdot b=17(17k+b) \Rightarrow k=\frac{b}{7b-17} \Rightarrow b\ge 7b-17 > 0 \Rightarrow b\in \emptyset.$$

Answer 2

$$a=\dfrac{17b}{7b-17}$$

Now if integer $d$ divides both $17b,7b-17;$

$d$ must divide $7(17b)-17(7b-17)=17^2$

So, $7b-17$ must divide $17^2$ to keep $a$ an integer

So, $7b-17$ must be one of $0,\pm1,\pm17,\pm17^2$ which is untenable.