$3 \cdot 60$ For which natural numbers $n$, is the number $3^{2 n+1}-2^{2 n+1}-6^{n}$ composite?
(24th All-Soviet Union Mathematical Olympiad, 1990)
[Solution]. The original expression can be transformed as follows:
$$
\begin{aligned}
& 3^{2 n+1}-2^{2 n+1}-6^{n} \\
= & 3^{2 n+1}-3^{n} \cdot 2^{n}-2^{2 n+1} \\
= & 3^{2 n+1}-3^{n+1} \cdot 2^{n}+3^{n} \cdot 2^{n+1}-2^{2 n+1} \\
= & 3^{n+1}\left(3^{n}-2^{n}\right)+2^{n+1}\left(3^{n}-2^{n}\right) \\
= & \left(3^{n}-2^{n}\right)\left(3^{n+1}+2^{n+1}\right) .
\end{aligned}
$$

When $n=1$, $3^{2 n+1}-2^{2 n+1}-6^{n}=13$ is a prime number.
When $n>1$, since
$$
3^{n}-2^{n}>1, \quad 3^{n+1}+2^{n+1}>1 \text{. }
$$

Therefore, $3^{2 n+1}-2^{2 n+1}-6^{n}$ is a composite number.
Thus, the required $n$ is a natural number where $n \geqslant 2$.