I'm interesting to know if there is general solution for the below equation and it's geometrical interpretation then my question here is :



> 
> 
> > 
> > **Question**
> > 
> > 
> > When is :$\displaystyle n!=x^n-y^n$ , with , $x,y,n$ are positive integers ?
> > 
> > 
> > 
> 
> 
> 



When is :$\displaystyle n!=x^n-y^n$ , with , $x,y,n$ are positive integers?
If $n=1$, the answer is yes. Take $x=2, y=1$ so that $1!=2-1$.


If $n>1$, the answer is no. Assume $d=\gcd(x,y)$ and $x=da, y=db, \gcd(a,b)=1$. Suppose 
$$n!=x^n-y^n=d^n(a^n-b^n). \tag1$$ 


Let $\nu\_p(z)$ denote the $p$-adic valuation of $z\in\mathbb{N}$ and $s\_p(n)$ be the sum of $p$-ary digits of $n$.


If a prime $p\,\vert\,a$ then $p$ does not divide $a^n-b^n$. Now, if $p\,\vert\,d$ then $\nu\_p(d^n)=n\cdot\nu\_p(d)>n$. But, Legendre's formula gives $\nu\_p(n!)=\frac{n-s\_p(n)}{p-1}<n$. That is, $\nu\_p(n!)) <\nu\_p(d^n(a^n-b^n))$. Absurd. Therefore, $p$ can not divide $d$ and hence $p$ can not divide $n!$ either (since (1) is assumed).


A similar analysis shows: if $p\vert b$ then $p$ does not divide $n$. Well, that means every prime divisor of $a$ is more than $n$. The same is true for $b$ unless $b=1$.


Since $a>b$ are both integers, it follows that $a\geq b+1$ and 
$$d^n(a^n-b^n)>(n+1)^n-n^n>(n+1)^{n-1}.$$
However, it's easy to see that $(n+1)^{n-1}>n!$ and hence a contradiction. So, (1) is impossible!


