That's not enough info!

Let $x=\overline{ab},y=\overline{ba}\\x=10a+b,y=10b+a$

Subbing in, we have

$99a^2-99b^2=m^2\\99|m\Rightarrow1089|m\\\mbox{Let }m=1089k^2,k\in\mathbb{N}.\\99a^2-99b^2=1089k^2\\a^2-b^2=11k^2\\(a-b)(a+b)=11k^2\\11|(a-b)\mbox{(rej)}\quad\mbox{or}\quad11|a+b\\\mbox{Clearly, }a+b<20\Rightarrow a+b=11\Rightarrow a-b=k^2$

Setting $k=1,2,3$ , we find that the unique solution is $a=6,b=5$ . Hence, we have $x+y+m=65+56+33=154$ .

Let x=10a+b Andy=10b+b Get x^2-y^2=m^2 That is 99(a+b)(a-b)=m^2. Here a and b are less than 10 positive integer. So a+b=11 and a-b =1 Hence x+y+m=154...