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Can you help me with this problem : Prove that for every positive integer $n>0$, $3\sum\limits_{i=1}^n i^5 $ is divisible by $\sum\limits_{i=1}^n i^3 $

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closed as off-topic by Travis, José Carlos Santos, John Omielan, Javi, YiFan Mar 27 at 21:41

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    $\begingroup$ proof by induction? $\endgroup$ – Yanko Mar 27 at 9:31
  • $\begingroup$ I would be interested to see an approach other than computing the closed forms. That would indicate a deeper connection. $\endgroup$ – robjohn Mar 27 at 15:38
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It is easy to see, when you look at the closed form of the sum. $$\sum_{j=1}^n j^3=\frac{1}{4} n^2(n+1)^2$$ $$\sum_{j=1}^n j^5=\frac{1}{12} n^2(n+1)^2(2n^2+2n-1)$$ If you divide, you will have a nice formula for the remainder.

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  • $\begingroup$ Thanks for the edit, i was on the phone. $\endgroup$ – Andrew Kovács Mar 27 at 9:56

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