Find all positive integers $w,x,y,$ and $z$ which satisfy $w! = x! + y! + z!$.
Good morning, Purdue!....
We have $w>x$,$w> y$ and $w>z$, obviously.
And $ w!-x!-y!-z! $ is minimal when $ x$, $y$, and $ z $ are maximal, relative to $w$.
Let´s consider then (suppose $ x$ , $ y$ and $ z$ different)
$(n+3)!-(n+2)!-(n+1)!-n!= n! [(n+3)(n+2)(n+1)-(n+2)(n+1)-1]$. This difference is always positive, never zero. So, in this conditions there are no $w$, $x$, $y$ and $z$ satisfying the equation.
If $x=y\neq z$, and $x>z$ we have (put $w=n+1$ and $x=n$) , $(n+1)!-2n!-z!= n!(n-1)-z!$, which is always positive (never zero).
If $x=y=z$, we have $(n+1)!=3n! \iff n=2$.
So $w=3$ and $x=y=z=2$.
Great proof, Luciano! BOILER UP!
There is only one such ordered-quadruple: (w,x,y,z) = (3,2,2,2).
For w < 3, w! = x! + y! + z! is impossible in the positive integers since 1! and 2! are both less than 3 (with x = y = z = 1 being the smallest possible combination on the RHS).
For w = 3, we obtain 3! = 2! + 2! + 2! (which is admissible).
For w > 3, the quantities x, y, and z cannot be more than w-1. Since w! = w(w-1)! and w > 3, there is no decomposition of w! into three smaller factorials.