, 2016

The only solution is $n = 7$.
Observe that
$$\begin{aligned} n &\equiv n \quad (\text{mod } 7), \\ n^2 + 10 &\equiv (n + 2)(n - 2) \quad (\text{mod } 7), \\ n^2 - 2 &\equiv (n + 3)(n - 3) \quad (\text{mod } 7), \\ n^3 + 6 &\equiv (n - 1)(n^2 + n + 1) \quad (\text{mod } 7), \\ n^5 + 36 &\equiv (n + 1)(n^4 - n^3 + n^2 - n + 1) \quad (\text{mod } 7). \end{aligned}$$
Since $n-3$, $n-2$, $n-1$, $n$, $n+1$, $n+2$ and $n+3$ are 7 consecutive integers, one of them must be a multiple of 7. It follows that one of the numbers is equal to 7, or otherwise it cannot be a prime. Clearly, $n^2 + 10 > 7$ and $n^5 + 36 > 7$.
If $n=7$, the 5 numbers are 7, 59, 47, 349, 16843, all of which are primes.