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.

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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.
