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 without loss of generality that $x\leq y$. We will necessarily have to have $y\leq z$, as well.
If $x=y$, then $x!+y!=2\cdot x!$, which can only be a factorial if $x=0$ or $x=1$. If $x<y$, then dividing by $x!$ yields $$1+y\cdot(y-1)\cdots(x+1)=z\cdot(z-1)\cdots(x+1).$$ We then clearly can't have $z=y$ (why?), so after subtraction and factoring, we have $$1=\bigl[z\cdots(y+1)-1\bigr]\cdot y\cdot(y-1)\cdots(x+1).$$ This only works if $$z\cdots(y+1)=2$$ and $$y\cdot(y-1)\cdots(x+1)=1.$$ (why?) This implies that $y=1=x+1$ and $z=2$.
Thus, you only need to consider triples $(x,y,z)$ of non-negative integers with $z=2$.
