Integer solutions of $x! = y! + z!$
$x! = y! + z!$ does not have any solutions in integers with $x \geq 3.$ As soon as $x \geq 3,$ we have $(x-1)! \leq x! / 3.$ With the necessary $y,z < x,$ we get $y! \leq x! / 3, \; \; z! \leq x! / 3,$ so $y! + z! \leq 2 x! / 3$ and $y! + z! \neq x!$