For the following equation, I've got only two solutions for x.
$x! + !x = a^3$, where $(x,a \in \Bbb W ) $
I've got x=1 and x=3.
Are there any other solutions? What methods can be used to find out solutions to problems like these without utilising the hit and trial method?
Note: The W stands for the set of all whole numbers. The 'x' and 'a' are both whole number variables. !x is the subfactorial function on x. http://en.wikipedia.org/wiki/Derangement