On a special divider obtained by permutation I want to find all positive integers $n$ and $m$ ($n > m$) verifying:
$\dfrac{n! + m}{m! + n} \in \mathbb{N}$
Note that there is a permutation of the factorial operator between the parts of the fraction.
Thanks for your help and suggestions !
 A: Since $m! + n > m$, a necessary condition is that $(n + m!) \not\mid n!$. That is a rather restrictive condition already.
We certainly have $(n + m!) \not\mid n!$ when $n + m!$ is prime, and if that is composite, it's not too common.
Since for composites other than $4$ and $9$ we have $N \mid \lfloor N/2\rfloor!$, we find that $n + m!$ must necessarily be prime if $n \geqslant m!$, because then $(n + m!)/2 \leqslant n$.
Now, to investigate the case where $p = n + m!$ is prime, we must have
$$(p - m!)! + m \equiv 0 \pmod{p}.\tag{1}$$
By Wilson's theorem, we have $(p-1)! \equiv -1 \pmod{p}$, so for $m = 1$ the condition is always fulfilled, and we have the family $(1,\,p-1)$ of solutions $(m,\,n)$, where $p > 2$ is a prime. For $m > 1$, the factorial $m!$ is even, hence
$$(p-m!)! \equiv - ((m!)!)^{-1}\pmod{p},$$
and plugging that into $(1)$ we obtain
$$m \cdot (m!)! \equiv 1 \pmod{p},\tag{2}$$
that is, $p$ must be a divisor of $m\cdot(m!)! - 1$. Since we demand $n > m$, we must exclude primes smaller than $m! + m$ from the list, if there are any.
For the first few very small $m > 1$, the complete list of such pairs is easy to obtain,
$$\begin{align}
m &= 2 & p = 3 & n = 1 \text{ too small}\\
m &= 3 & p \in \{17,\, 127\} & n \in \{11,\, 121\}\\
m &= 4 & p \in \{71,4656788681,7506211238849\} & n \in \{47,4656788657,7506211238825\}
\end{align}$$
For $m = 5$, we get $7347398537599573$ and a rather large prime, 184 digits. For $m = 6$, I only found the prime factor $11527$ of $6\cdot 720! - 1$ so far, there are more, but they are too large for me to find.
For $m! + n$ composite, the only solution I found so far is $m = 4$ and $n = 14$.
A: Small idea but not much advance:
We can first write $n=m+k$ and get $$\frac{n!+m}{m!+n}=\frac{(m+k)!+m}{m!+(m+k)}.$$
Now we divide a little 
\begin{align}
\frac{(m+k)!+m}{m!+(m+k)}&=\frac{m!(m+1)\dotsm(m+k)+m}{m!+(m+k)}\\&=(m+1)\dotsm(m+k)+\frac{m-(m+1)\dotsm(m+k-1)(m+k)^2}{m!+(m+k)}.
\end{align}
So the problem reduces to find when $$\frac{m-(m+1)\dotsm(m+k-1)(m+k)^2}{m!+(m+k)}$$ is an integer. The advantage is that now the numerator is a polynomial in $m$, which grows slower than the $m!$ in the denominator.
