Mathematical Reflections 2012, Issue 4 - Problem O235

Problem:
Solve in integers the equation $$xy-7\sqrt{x^2+y^2}=1.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $xy \geq 1$. By symmetry, we reduce to $x,y > 0$. Rewriting the equation in the form $(xy-1)^2=49(x^2+y^2)$, we put $xy=7t+1, t \geq 0$, so that $x^2+y^2=t^2$. Then $$(x+y)^2=x^2+y^2+2xy=t^2+2(7t+1)=(t+7)^2-47,$$ which gives $$(t+7-x-y)(t+7+x+y)=47.$$
So, we must solve the systems
$$\left\{\begin{array}{rcl} t+7-x-y & = & 1 \\ t+7+x+y & = & 47 \end{array} \right. \qquad \left\{\begin{array}{rcl} t+7-x-y & = & -47 \\ t+7+x+y & = & -1 \end{array} \right.$$
Solving the first, we get $t=17$ and $x+y=23$, solving the second we get $t=-31<0$, i.e. no solution. So we have $x+y=23, xy=120$, which gives $x=15,y=8$ and $x=8, y=15$. In conclusion, the solutions are $$(15,8),(8,15),(-15,-8),(-8,-15).$$