KVPY #2

Let $x = \overline{ab}$ where $9 \ge a \gt b \ge 1$ . Then $x = 10a + b$ and $y = 10b + a$ , and so

$x^{2} - y^{2} = (10a + b)^{2} - (10b + a)^{2} = 100a^{2} + 20ab + b^{2} - (100b^{2} + 20ab + a^{2}) =$

$99(a^{2} - b^{2}) = 3^{2} \times 11 \times (a + b)(a - b)$ .

For this to be a perfect square we will require that $(a + b)(a - b) = 11n^{2}$ for some positive integer $n$ . For this to be the case, since $11$ is prime and $a - b \lt 11$ , we must have $11|(a + b)$ . But as $a + b \lt 22$ we must have $a + b = 11$ , in which case the possible options for $(a,b)$ are $(9,2), (8,3), (7,4)$ and $(6,5)$ . Finally, given that $a + b = 11$ we will require that $a - b = n^{2}$ for some positive integer $n$ , which of the possible options is only achieved when $(a,b) = (6,5)$ .

Thus $x = 65, y = 56$ and $x^{2} - y^{2} = 99(6 + 5)(6 - 5) = 3^{2} \times 11^{2} = 33^{2}$ , and so

$x + y + m = 65 + 56 + 33 = \boxed{154}$ .