Find all solutions of the equation $x! + y! = z!$ Not sure where to start with this one.  Do we look at two cases where $x<y$ and where $x>y$ and then show that the smaller number will have the same values of the greater?  What do you think?

Suppose that $0 \le x \le y$ WLOG. Then $y! < x! + y! \le 2(y!) < (y+1)!$ for $y \ge 2$, so no solutions with $y \ge 2$ are possible (because then every $x! + y!$ is between two consecutive factorials and not equal to either one).
