Proof that $(3^n - 2^n) / n$ is not an integer, $n \geq 2$ Trying to prove that $(3^n - 2^n)/n$ is not an integer for $n\geq 2$.

Was trying something along the lines of induction with:

$3^{n+1} - 2^{n+1} = 2(3^n - 2^n) + 3^n \equiv 0 \mod (n+1)$

But that gets messy...
Let $p$ be the smallest prime factor of $n$ and write $n = p^k m$ where $p \nmid m$. By repeated application of Fermat's little theorem it follows that $3^n - 2^n \equiv 3^m - 2^m \equiv 0 \bmod p$. If $p = 2, 3$ then this is clearly never possible; otherwise, it follows that $\left( \frac{2}{3} \right)^m \equiv 1 \bmod p$, hence that $\gcd(p-1, m) &gt; 1$. But this is also impossible since the prime factors of $m$ are all larger than $p$.
I would consider first the case that $n \geq 3$ is prime.  Using Fermat's Little Theorem, you should be able to show $3^{n} - 2^{n} \equiv 1 \text{ (mod }n)$.

EDIT:  I retract what I said about the composite case.  I'll keep thinking about it.