Given the natural numbers x and y satisfy (1 + x!)(1 + y!) = (x + y)! What is the value of x + y?

Question

x and y are distinct 2 digit numbers such that y is obtained by reversing the digits of x.
Suppose they also satisfy x^2 - y^2 = m^2 for some positive integer m, then find the value of x+y+m?

Answer 1

I think (x + y) = 3 as x and y can only take the values 1 and 2 and any other values of x and y will either lead to fractions or 0.