TITLE: Finding all the numbers that fit $x! + y! = z!$
QUESTION [16 upvotes]: I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon  inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it. 
From the definition of the factorial function we know $x! = x(x-1)(x-2)...(2)(1)$
So we can do something like this:
$$ [x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)]$$ 
we can then factor all of the common terms out on the LHS. 
$$ [...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)]$$
and divide the common terms out of the right hand side 
$$[x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...]$$
I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution). 
If anybody can provide a hint as to how to proceed I would appreciate it.

REPLY [4 votes]: If $x>1$ and $ y \leq x$ then
$$x!<x!+y! \leq 2x!<(x+1)!$$
therefore
$$x!<z!<(x+1)!$$
so that the only solutions are $(x,y,z)=(0,0,2),(0,1,2),(1,0,2),(1,1,2)$.