Prove that there is no integer $n \geq 2$ for which $$\frac{3^n - 2^n}{n}$$ is an integer
I really don't know how to start with it except with the parity of n (n being even clearly doesn't make the fraction an integer)
Edit: If I take n=pk for a prime p then all i get is to disprove the congruence 3^k=2^k (mod p) which I don't know how to proceed with
Hint : try to pick Prime $p>3$ s.t. $n = pk$ and $\gcd(p-1, n) = 1$
The existence is below
pick the smallest prime divisor $p$ of n, since $2, 3 \nmid 3^n - 2^n$, so $p>3$
More steps and why pick $\gcd(p-1, n) = 1$ :
Let $a = 2^{-1}3 \pmod{p}$, by $a^{p-1} = 1 \pmod{p}$, $a^n = 1 \pmod{p}$, we can obtain $a = a^{\gcd(p-1, n)} = 1 \pmod{p}$, but $2^{-1}3 \neq 1 \pmod{p}$.
Suppose prime $\,\color{#0a0}{p\mid n\mid 3^n\!-2^n}.\,$ Then $p$ is odd so $\bmod p\!:\ 2$ is invertible so $\,3/2\,$ exists so
$$ \bmod p\!:\ \ (3/2)^{\large p-1}\!\equiv \color{#0a0}{1\equiv (3/2)^{\large n}}\Rightarrow\, (3/2)^{\large \color{#c00}{(p-1,n)}}\equiv 1$$
However if $\,p\,$ is the least prime factor of $\,n\,$ then $\,\color{#c00}{(p-1,n) = 1}\ $ therefore
$$ 1 \equiv (3/2)^{\large \color{#c00}{(p-1,n)}}\!\equiv 3/2\,\Rightarrow\,2\equiv 3\,\Rightarrow\,p\mid 1\Rightarrow\!\Leftarrow$$
Remark $\ $ The key idea is: $ $ if $\, a^n\! = 1,\ a\neq 1\,$ then $\:a\:$ has order $\,\ge\,$ the least prime factor $\,p\mid n.\,$ In particular this implies that $\rm\ a^{p-1}\!\ne 1,\, $ as above.
This is a slight variant of the proof I gave in this answer. See there for generalizations.
3^k=2^k (mod p) which I don't know how to proceed with
– SuperMage1 Jun 08 '17 at 14:39