Let $x,y$ be nonnegative integers satisfying $(xy-7)^2 = x^2+y^2$. Prove that $x+y = 7$ 
Let $x,y$ be nonnegative integers satisfying $(xy-7)^2 = x^2+y^2$. Prove that $x+y = 7$.

We can rewrite the given equation as $$(x+y)^2-(xy-6)^2 = (x+y-xy+6)(x+y+xy-6) = 13.$$ How do we continue?

Note that we now have that either $$x+y-xy+6 = \pm 1 \tag{1}$$ 
$$x+y+xy-6=\pm 13 \tag{2}$$
or $$x+y-xy+6= \pm 13 \tag{3}$$ $$x+y+xy-6= \pm 1 \tag{4}$$
From your last equation. $\frac{(1)+(2)}{2}$ and $\frac{(3)+(4)}{2}$ both give that $$x+y=\pm 7$$From this, using the condition that $x$ and $y$ are non negative integers, we see that $x+y=7$.
