FOM - 1

Suppose $x \ge y$ for now. Then $\begin{aligned} x^2+y^2 &\le 2x^2 \\ (xy-7)^2 &\le 2x^2 \\ xy-7 &\le x\sqrt{2} \\ x &\le \frac7{y-\sqrt{2}} \end{aligned}$ Now if $y \ge 4$ this gives $x \le \frac7{4-\sqrt{2}} < 4$ , which is a contradiction, so we only have to check $y= 0,1,2,3$ .

For $y = 0$ we get $x = 7$ . For $y = 1$ we get $x = 24/7$ , not an integer. For $y = 2$ we get $3x^2-28x+45 = 0$ , and it's easy to check that the solutions are not integers. For $y = 3$ we get $8x^2-42x+40 = 0$ , which factors as $2(4x-5)(x-4) = 0$ . So $x = 4$ is the unique integer solution in that case.

Hence there are two solutions with $x \ge y$ , $(7,0)$ and $(4,3)$ . The equation is symmetric, so there are four solutions in all, $(7,0),(4,3),(3,4),(0,7)$ . The sum of the $x$ -values is $\fbox{14}$ .

$xy - 7 = \sqrt{x^2 + y^2} \leq x + y \\ xy - 7 \leq x + y \\ x + y - xy + 7 \leq 0 \\ (x-1)(1-y) + 8 \leq 0 \\ (x-1)(y-1) \leq 8$

A pretty tight bound for our variables considering the little amount of work we had to put into it. Let's make it tighter by trying small values of $y$ .

$y = 0 \implies 7^2 = y^2 \implies y=7 \\ y=1 \implies (x-7)^2 = x^2 + 1 \implies x^2 - 14x + 49 = x^2 + 1 \implies 14x = 48 \implies x \not \in \mathbb{Z} \\ y = 2 \implies (2x - 7)^2 = x^2 + 4 \implies 4x^2 - 28x + 49 = x^2 + 4 \implies 3x^2 - 28x + 45 = 0 \implies x \not \in \mathbb{Z}$

This last assertion is proven by considering the discriminant of the polynomial which evaluates to 244 which is not a perfect square. We now test $y=3$ which results in $8x^2 - 42x + 40 = 0 \implies x = 4$ . Now any other solution would imply that $y \geq 4 \implies y-1 \geq 3 \implies x-1 \leq \frac{8}{3} \implies x \leq 3$ . We no longer need to check anymore because the equation is symmetric so finding solutions such that $x \leq 3$ is equivalent to finding solutions such that $y \leq 3$ which is what we just did. Thus, the possible values are $0,7,3,4$ .

Alternative: Another way of using the symmetry of the equation to realize we only need to check $x \leq 3$ is by assuming that $y \geq x$ which would imply $(x-1)(x-1) \leq 8 \implies (x-1)^2 \leq 8 \implies x-1 \leq \sqrt{8} \implies x \leq \sqrt{8} + 1 \approx 3.8284 \implies x \leq 3$ .

$x$ can be $0,3,4,7$ Part of this problem involves the methods of generating a $primitive$ Pythagorean triple. Secondly, keep in mind that $x$ and $y$ are non negative integers; $0$ will work.