imc

Let $x=10a+b,y=10b+a$. The given conditions imply $x>y$, which implies $a>b$, and they also imply that both $a$ and $b$ are nonzero. Then, $x^2-y^2=(x-y)(x+y)=(9a-9b)(11a+11b)=99(a-b)(a+b)=m^2$. Since this must be a perfect square, all the exponents in its prime factorization must be even. $99$ factorizes into $3^2\cdot 11$, so $11|(a-b)(a+b)$. However, the maximum value of $a-b$ is $9-1=8$, so $11|a+b$. The maximum value of $a+b$ is $9+8=17$, so $a+b=11$. Then, we have $33^2(a-b)=m^2$, so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$. We know $a+b=11$, and, for $a-b=1$, adding equations to eliminate $b$ gives us $2a=12⟹a=6,b=5$. Testing $a-b=4$ gives us $2a=15⟹a=\frac{15}{2},b=\frac{7}{2}$, which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b)=(6,5)$, and $x+y+m=\boxed{154}$.