Clearly true if n is a multiple of 2 or 3. Also, you can show it's true for n prime using Fermat's little theorem. You can also show that if n divides 3{a+b} - 2^{a+b} and n divides 3^a - 2^a then n divides 3^b - 2^b.


Not sure where to go from there though, or if I should be trying something completely different...



Trying to solve: Show that n does not divide 3^n - 2^n for n greater than or equal to 2.
Take $p$ to be the smallest prime divisor of $n$. You have that $p$ divides $3^{p-1}-2^{p-1}$ and also $3^n-2^n$. So $p$ divides $3^{\operatorname{gcd}(p-1,n)}-2^{\operatorname{gcd}(p-1,n)}$. However it is easy to see that this gcd must equal $1$, so $p$ divides $3-2$, and we obtain the desired contradiction.


