A bit of progress/motivation: Primes relatively prime to 30 ^4, euler totient formula for 30 is 8, so prime^4=+-1 mod 30. Not sure how to continue to show three CONSECUTIVE primes though.

Set $S=\textstyle \sum_{1\le i\le 31}n_i^4$. Note that if $5\notin \{n_1,\dots,n_{31}\}$ then $n_i^4\equiv 1\pmod{5}$ by Fermat, yielding $S\equiv 31\equiv 1\pmod{5}$, a contradiction. So, $5\in \{n_1,\dots,n_{31}\}$. Similarly, if $n_i$ are all odd, then $S$ is clearly odd, another contradiction. Lastly, if $3\notin\{n_1,\dots,n_{31}\}$ then $S\equiv 31\equiv 1\pmod{3}$. Thus, $2,3,5\in\{n_1,\dots,n_{31}\}$, as desired.