A Rocket of Primes

Obviously, $n \neq 1$ . If $n = 2$ , $2n^2 + 5n + 8$ and $n^2 + 398n +12$ will be composite. If $n = 3$ , $12n + 13$ and $n^2 + 398n + 2$ will be composite. Since $4$ is composite, $n \neq 4$ .

Checking for $n = 5$ ,

$n^2 + 4 = 29$

$2n^2 + 5n + 8 = 83$

$12n + 13 = 73$

$24n - 11 = 109$

$n^2 + 398n + 12 = 2017$

which are all primes! Hence $n=5$ .

To prove that there is no other solution for $n>5$ , set $n$ as $5k, 5k+1, 5k+2, 5k+3,$ and $5k+4$ (for all $k \geq 1$ ). By substituting, there is always at least one among the numbers which is divisible by $5$ (and perhaps, other prime numbers).

Hence, $\boxed{n=5}$