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 $x$.

Question

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.

Answer 1

$$(xy-7)^2=x^2+y^2 $$ $$x^2y^2-14xy+49 = x^2+y^2 $$ $$x^2y^2-12xy+49 = x^2+2xy+y^2 $$ $$(xy-6)^2+13=(x+y)^2 $$ $$13 = (x+y)^2-(xy-6)^2 $$ $$13 = (x+y+xy-6)(x+y-xy+6) $$ The factors on the right must be factors of the prime $13$. We conclude $x+y=7$ and easily enumerate all solutions.

Answer 2

Write the equation as $(xy-7)^2 - x^2 = y^2$, and factor as $$(xy+x-7)(xy-x-7) = y^2$$

Note that if $x \ge 4$ and $y \ge 4$, $$xy+x-7 > xy - x - 7 \ge 4 (y-1) - 7 = y + 3y - 11 > y$$

Now look at small values of $x$ or $y$.

Answer 3

$$(x+xy-7)(x-xy+7) + y^2 = 0$$

$$(x(1+y)-7)(x(1-y)+7) + y^2 = 0$$

This means for example prime factors of $y^2$ must split over factors of $$(7-x(y+1))(7+x(1-y))$$

They will also be opposite (mod $7$) and same (mod $x$).

But we can start with that for each side be positive ( requirement for a square ) both factors of a product must be either negative or positive.

1. both negative: $\cases{7<x(y+1) \\ 7<-x(1-y) = x(y-1)}$

2. both positive $\cases{7>x(y+1) \\ 7>x(y-1)}$

Now this should reduce the problem enough to check the rest by hand.

Answer 4

We have $$(xy)^2-12xy+49=(x+y)^2\iff(xy-6)^2+13=(x+y)^2$$ hence $$13=(x+y+xy-6)(x+y-xy+6)$$ It follows the only possibility (after discard$-13$ and $-1$) $$x+y+xy-6=13\\x+y-xy+6=1$$ 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 $(x,y)=(0,7),(7,0),(4,3),(3,4)$ $$S=0+7+4+3=\color{red}{14}$$