Fall 2017, problem 49

Good morning, Purdue!....

Answer: 3,2,2,2.

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 $x=y=z=2$.

There is only one such ordered-quadruple: $(w,x,y,z) = (3,2,2,2)$.

PROOF:

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.

QED