Matematické Fórum

stuart clark · 27. 10. 2011 15:51

find value of $x\;,y$ and $z$ in $x!+y!=z!$

Where $x\;,y\;,z\in\mathbb{Z^{+}}$

and $n!=n\times (n-1)\times (n-2)\times ..........\times 3\times 2\times 1$

vanok · 27. 10. 2011 17:54 — Editoval vanok (27. 10. 2011 18:39)

Hi ↑ stuart clark:,
I use the following reasoning:
Without losing the majority we can suppose that $x \le y \le z$

First case: $0<x<y$
Let us suppose that
$x!+y!=z!$
is true.
The division by $y!$ of this equality gives:
$\frac{x!}{y!}=1 -\frac{z!}{y!}$
This is a contradiction, because to the left we have a fraction and to the right an integer.

Second case $x = y$
The division by $x!$ of our equality gives:
$\frac{z!}{x!}=2$
This gives 2 trivial solutions x=y=0 and x=y=1