# Prove that for every positive integer $n>0$, $3\sum\limits_{i=1}^n i^5$ is divisible by $\sum\limits_{i=1}^n i^3$ [closed]

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 $$\sum\limits_{i=1}^n i^3$$

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

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• proof by induction? – Yanko Mar 27 at 9:31
• I would be interested to see an approach other than computing the closed forms. That would indicate a deeper connection. – robjohn Mar 27 at 15:38

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.