Let $p$ be a prime which is congruent to $1\bmod 4$ and $p\neq 17$. We will show that $n=p$ satisfies the condition. Since $n-2$ is odd and $4\mid n-1$, we have $16\mid ( n-2)^{n-1} -1$ So we have $17\mid 3^{16} -1\mid $$3^{( n-2)^{n-1} -1} -1$. Note that $( n-2)^{n-1} -1\equiv ( -1)^{n-1} -1\equiv 0\pmod{p-1}$. Note that $( n-2)^{n-1} -1\equiv 1-1\equiv 0\pmod p$. So we have $p^{2} \mid 3^{p( p-1)} -1\mid 3^{( n-2)^{n-1} -1} -1$. Hence we have $17p^{2} \mid 3^{( n-2)^{n-1} -1} -1$ as desired. $\blacksquare $