If $30$ divides $p_1^4 + p_2^4 + \ldots + p_{31}^4$. Prove that $p_1=2$, $p_2=3$ and $p_3=5$. Let $p_1<p_2<\cdots<p_{31}$ be prime numbers such that $30$ divides $p_1^4 + p_2^4 + \cdots + p_{31}^4$. Prove that $p_1=2$, $p_2=3$ and $p_3=5$.
No clue how to start..Hints are welcomed. 

First, if $p_1\neq 2$ then $p_1>2$ and so $p_1$, $p_2$, $\ldots$, $p_{31}$ are all odd, in such a case $p_1^4+p_2^4+_\ldots+p_{31}^4$ is odd, in particular 30 doesn't divide $p_1^4+p_2^4+_\ldots+p_{31}^4$. This shows $p_1=2$.
If $p_2\neq 3$ then $p_2\equiv \pm 1(\mod 3)$ and so $p_i^{4}\equiv 1(\mod 3)$ for $i=1,2,\ldots,31$, hence $p_1^4+p_2^4+_\ldots+p_{31}^4\equiv 1 (\mod 3)$ that means $p_1^4+p_2^4+_\ldots+p_{31}^4$ is not a multiple of $30$. So $p_3=3$. A similar argue show us that $p_3$ should be $5$: $p_3\neq 5 \Longrightarrow p_i^4\equiv 1 (\mod 5)$ for $i=1,2,\ldots,31$ by Little Fermat Theorem and $p_1^4+p_2^4+_\ldots+p_{31}^4\equiv 1 (\mod 5)$.
