If $30$ divides $p_1^4 + p_2^4 + \ldots + p_{31}^4$. Prove that $p_1=2$, $p_2=3$ and $p_3=5$.

Question

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.

Answer 1

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 $p_1^4+p_2^4+_\ldots+p_{31}^4\equiv 1 (\mod 5)$.

Answer 2

Hint from comment - look at parity for $2$ and use Little Fermat.

Answer 3

Let $a=1$ if $2$ is among the $p_i$'s, and $a=0$ else. Define similarly $b$ for $3$ and $c$ for $5.$

For any prime number $p$ coprime to $30,$ we have $p^4 ≡ 1\bmod2,3,5$ hence $\bmod{30}.$ Therefore, the sum of the fourth powers of the $p_i$'s different from $2,3,5$ is congruent to the number of these $p_i$'s, i.e. to $31-a-b-c.$

Now, $\bmod{30},$ $$2^4\equiv-14,\quad3^4\equiv-9,\quad5^4\equiv-5.$$ Alltogether, $$30\mid-14a-9b-5c+31-a-b-c=31-15a-10b-6c.$$ Since $a,b,c\in\{0,1\},$ the only possibility is $a=b=c=1.$