Find all positive integer $n$ such that for all prime $p$: $$\frac{pn}{p+n} \in \mathbb{Z}$$
I have tried these:
Case 1:$n$ odd
If choose $p=2$ will have: $$\frac{2n}{2+n}=\frac{2(2+n)-4}{2+n}=2-\frac{4}{2+n}$$
So: $(2+n)|4$ which is wrong because $n$ is odd
Case 2:$n$ even
In this case, I can't do anything!
Suppose $n>0$ and $p+n$ divides $pn$. If $n\wedge p=1$ then $p\wedge (p+n)=1$ hence $p+n$ divides $pn$ entails by the Gaus lemma that $p+n$ divides $n$ which is impossible as $p+n>n>0$. So $n$ must be $\lambda p$ for an integer $\lambda>0$ and as $n+p=(\lambda+1)p$ we have $\frac{np}{n+p}=\frac{\lambda p}{\lambda+1}$ and using $\lambda\wedge (\lambda+1)=1$ and the Gauss lemma we must have then that $1+\lambda$ divides $p$, which entails $\lambda=0$ or $\lambda=p-1$. Thus for any prime $p$ the only strictly positive integer $n$ such that $p+n$ divides $pn$ is $(p-1)p$
If $\frac{pn}{p + n}$ is an integer for all primes $p$, then so is $n - \frac{pn}{p + n} = \frac{n^2}{p + n}$ for all primes $p$.
But by the infinitude of primes, $p$ can be arbitrarily large, so there exists some $p$ with $p + n > n^2$. Hence, $\frac{n^2}{p + n}$ won't be an integer (since $n^2 > 0$).
We conclude there are no solutions.
