4. $4 \star$ For which natural numbers $n$ is $3^{2 n+1}-2^{2 n+1}-6^{n}$ composite?

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When $n>1$, $3^{2 n+1}-2^{2 n+1}-6^{n}=\left(3^{n}-2^{n}\right)\left(3^{n+1}+2^{n+1}\right)$, and $3^{n}-2^{n}>$ 1, so when $n>1$, $3^{2 n+1}-2^{2 n+1}-6^{n}$ is a composite number. While when $n=1$, $3^{2 n+1}-2^{2 n+1}-6^{n}=13$ is a prime number.