2. Determine all natural numbers $n$ for which $3^{2 n+1}-2^{2 n+1}-6^{n}$ is a prime number.
Solution. We have

$$
3^{2 n+1}-2^{2 n+1}-6^{n}=3 \cdot 3^{2 n}-2 \cdot 2^{2 n}-3^{n} \cdot 2^{n}=3 \cdot\left(3^{n}\right)^{2}-2 \cdot\left(2^{n}\right)^{2}-3^{n} \cdot 2^{n}
$$

We introduce the substitution $3^{n}=a$ and $2^{n}=b$, and we get

$$
3 a^{2}-2 b^{2}-a b=3 a^{2}-3 a b+2 a b-2 b^{2}=3 a(a-b)+2 b(a-b)=(a-b)(3 a+2 b)
$$

Thus, $3^{2 n+1}-2^{2 n+1}-6^{n}=\left(3^{n}-2^{n}\right)\left(3^{n+1}+2^{n+1}\right)$. For $n \geq 2$, we have $3^{n}-2^{n}>1$, so $\left(3^{n}-2^{n}\right)\left(3^{n+1}+2^{n+1}\right)$ is a composite number. For $n=1$, we have $(3-2)\left(3^{2}+2^{2}\right)=13$, which means that $3^{2 n+1}-2^{2 n+1}-6^{n}$ is a prime number only for $n=1$.