I was checking the following number theory exercise:
Show that:
If $1 ≤ x ≤ y ≤ z$ and $x! + y! = z!$ , then $x = y = 1$ and $z = 2$. (Hint: Use congruences $mod$ $y!$)
Where should I apply the congruences $mod$ $y!$? I don't know how to work with variable factorials in modular arithmetic, any help will be really appreciated.
Since for $y\le z$ we have $y!|z!$ we obtain$$y!|z!-y!=x!$$which yields to $y\le x$ and therefore $x=y$, so we have$$z!=2\cdot x!$$if $z\ge x+2$ we have $$z!>(x+1)!\ge 2x!$$which is impossible. Also $z\ne x$ therefore we have $z=x+1$. By substitution we finally obtain$$(x+1)!=2x!$$or$$x=y=1\\z=2$$