2005 AMC 12B Problem 19

Let $x = 10a + b$ and $y = 10b + a$ with $a \gt b.$ Then $$\begin{aligned} &x^2 - y^2 = (10a+b)^2 \\ &\quad {}- (10b+a)^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned}$$ Since $99 = 9 \cdot 11,$ for this to be a perfect square we need $(a+b)(a-b)$ to be a multiple of $11.$ As $a + b \le 17$ and $a - b \le 8,$ the only multiple of $11$ available is $a + b = 11.$ Then $x^2 - y^2 = 9 \cdot 11^2 (a - b),$ which is a perfect square exactly when $a - b$ is a perfect square. Because $a+b=11$ is odd, $a-b$ is odd; and because $1 \le a-b \le 8,$ its only possible square value is $1.$ Hence $(a, b) = (6, 5).$ So $x = 65,$ $y = 56,$ and $m = \sqrt{65^2 - 56^2}$ $= \sqrt{1089} = 33.$ Thus $x + y + m$ $= 65 + 56 + 33 = 154.$ Thus, the correct answer is E.