Find all values of $n \in \mathbb{N}$ such that $3^{2 n}-2^{n}$ is prime.
The expression we are discussing in terms of primality can be rewritten as $3^{2 n}-2^{n}=\left(3^{2}\right)^{n}-2^{n}=9^{n}-2^{n}=(9-2)\left(9^{n-1}+9^{n-2} \cdot 2+9^{n-3} \cdot 2^{2}+\cdots+2^{n-1}\right)$.

For $n \geqslant 2$, the right factor is at least $9^{1}$, so it is different from 1. The left factor is itself different from 1, so we conclude that $3^{2 n}-2^{n}$ is composite.

It suffices to handle the remaining cases by hand: for $n=1,9-2=7$ is prime, for $n=0$, $0-0=0$ is not prime.

The set we are looking for is therefore $\{1\}$.

Remark 1.1. Factoring an expression $n$ as $n=a b$ is a very classic method to show that a number is not prime. But be careful, the factorization represents only half of the work. It remains to verify that the two factors are strict divisors. This can be done by checking that $a>1$ and $b>1$ as we did earlier, or by checking that $1<a<n$, or even by checking that $\mathrm{a}<\mathrm{n}$ and $\mathrm{b}<\mathrm{n}$.