1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Parity for $2$ and little Fermat? $\endgroup$ – Mark Bennet May 5 '14 at 13:16
  • $\begingroup$ @MarkBennet I think you have solved the problem; why not post it as an answer? $\endgroup$ – awllower May 5 '14 at 13:19
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks for the hints and answers. $\endgroup$ – Morty May 5 '14 at 13:35
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.