Finding all the numbers that fit $x! + y! = z!$

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.

Solution 1:

If $x, y \in \{0,1\}$, then we can always find a solution $z \in \{0, 1, 2\}$. The rest of the post will show that there are no other solutions.

Let us assume $y \geq x \geq 2$ without loss of generality.

Dividing both sides by $x!$ gives $$ 1 + y(y-1)\cdots(x+1) = z(z-1)\cdots(x+1). $$ If $y > x$, we see $x+1$ divides the right-hand side but not the left-hand side ($x+1$ divides one term in the sum but not the other), in which case there are no solutions.

If $y = x$, we may reduce the problem to that of solving $2y! = z!$. Since $y \geq 2$, the left-hand side always has more factors of $2$ than the right-hand side, in which case there are no solutions.

Solution 2:

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 $(x,y,z)=(0,0,2),(0,1,2),(1,0,2),(1,1,2)$.

Solution 3:

Very shortly, $(x,y,z)=(0,0,2),(0,1,2),(1,0,2),(1,1,2)$ are the only solutions because if WLOG $1<x\le y<z$ then dividing by $y!$ yields $$1+\frac{1}{(x+1)(x+2)\cdots y}=(y+1)(y+2)\cdots z,$$ whose RHS is an integer while its LHS is not.