<s>[INST] $30\mid p_1^4+p_2^4+\cdots+p_{31}^4$. Prove that $p_1=2,p_2=3,p_3=5$ Let $p_1<p_2<\cdots<p_{31}$ be prime numbers such that $30\mid p_1^4+p_2^4+\cdots+p_{31}^4$. Prove that $p_1=2,p_2=3,p_3=5$
I have no clue whatsoever about how to even approach this problem. Any hint is welcome.
 [/INST] There are $30$ terms in $p_2^4+...+p_{31}^4$ and each of them is $\equiv 1\pmod 2.$ So $2$ divides $p_1^4+30$ so $2$ divides $p_1^4$ so $p_1=2$.  
There are $29$ terms in $p_3^4+...+p_{31}^4 $  and each of them is $\equiv 1 \pmod 3 .$  So $3$ divides $2^4+p_2^4 +29$ so $3$ divides $p_2^4$ so $p_2=3.$
There are $28$ terms in  $p_4^4+...+p_{31}^4$ and each of them is $\equiv 1 \pmod 5 .$  So $5$ divides $2^4+3^4+p_3^4+28,$ so $5$ divides $p_3^4$ so $p_3=5.$ 
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