If $x$ and $y$ are non-negative integers for which $(xy-7)^2=x^2+y^2$. Find the sum of all possible values of $x$. I am not able to reach to the answer. I have used discriminant as $x$ and $y$ are both integers but it didn't give any hint to reach to answer. I am not able to understand how should I deal with these type of question. 

$$(xy)^2−12xy+49=(x+y)^2\longleftrightarrow(xy−6)^2+13=(x+y)^2.$$ Hence $$13=(x+y+xy−6)(x+y−xy+6).$$ It follows the only possibility (after discarding $−13$ and $−1$) is that $$x+y+xy−6=13,$$ $$x+y−xy+6=1.$$ This implies clearly $x+y=7$ which gives the candidates $(0,7),(1,6),(2,5),(3,4)$ and symmetrics, and it is verified that the only to be accepted are $(0,7)$ and $(4,3)$.
Consequently, taking $(x,y)=(0,7),(7,0),(4,3),(3,4)$, $$S=0+7+4+3=14.$$
