# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, José Carlos Santos, John Omielan, Javi, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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

## 1 Answer

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.

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