Proof that if $3^n - 2^n$ is prime then $n$ is prime

This was a question from a previous year in a test and I couldn't solve it yet.

If $3^n - 2^n$ is prime, then $n$ must be prime.

Do you have any tips, suggestions?

• If $a\mid b$, then $3^a-2^a$ is a divisor of $3^b-2^b$. Jan 8 '17 at 18:26

Suppose that $n$ is composite, say $n = ab$. Then $$3^n - 2^n = (3^a)^b - (2^a)^b,$$ and $$x^b - y^b = (x -y) (x^{b-1} + x^{b-2}y + \cdots + y^{b-1}).$$ Taking $x = 3^a$, $y = 2^a$ gives a non-trivial factorization of $3^n - 2^n$. Hence if $n$ is composite, then $3^n - 2^n$ is as well.