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?

Let $x \le y < z$:
We get $x! = 1\cdot 2 \cdots x$, $y! = 1\cdot 2 \cdots y$, and $z! = 1\cdot 2 \cdots z$.
Now we can divide by $x!$
$$1 + (x+1)\cdot (x+2)\cdots y = (x+1)\cdot (x+2) \cdots z$$
You can easily show by induction that the right side is bigger than the left for all $z>2$. The only cases that remain are $0! + 0! = 2!$, $0! + 1! = 2!$, $1! + 0! = 2!$, and $1! + 1! = 2!$.
