2005 AMC 10B Problem 24

Write $x = 10a + b$ and $y = 10b + a$ with $a \gt b.$ Then $$\begin{aligned} m^2 &= x^2 - y^2 \\ &= 99(a^2 - b^2) \\ &= 99(a+b)(a-b). \end{aligned}$$ Since $99 = 9 \cdot 11,$ for $m^2$ to be a perfect square we need $(a+b)(a-b)$ to be divisible by $11.$ As $a + b \le 17,$ this forces $a + b = 11,$ and then $a - b$ must itself be a perfect square. With $a - b \le 8,$ the only workable case is $a - b = 1,$ giving $(a, b) = (6, 5).$ Then $x = 65,$ $y = 56,$ and $m^2 = 99 \cdot 11 = 33^2,$ so $m = 33.$ Therefore $x + y + m$ $= 65 + 56 + 33$ $= 154.$ Thus, E is the correct answer.