I tried to prove it by simple binomial expansion but I am stuck. I assumed, $n=(2k+1) ,k=1,2,3,......$ and after expanding the expression $(2k+1)^5-(2k+1)$ I arrived at this expression$→$ $8k+40k^2+80k^3+80k^4+32k^5$. Now I cannot proceed further with the first, second and last term hindering my approach..
Can anyone please help me with this? Also, it will be great if you can give me some other ideas (like with Fermat's theorem) for this sum. Thank you..
$n^5-n$ is divisible by $5$ by Fermat's little theorem.
For $n$ odd, $n^2-1$ is divisible by $8$ and $n^2+1$ is divisible by $2$,
so $n^5-n=n(n^2-1)(n^2+1)$ is divisible by $16$.
Since $n^5-n$ is divisible by $5$ and $16$, it is divisible by $80$.
Little fermat implies that $n^4-1=0$ mod $5$, $(2k+1)^2=4k^2+4k+1=4k(k+1)+1$, remark that $k(k+1)$ is even implies that $4k^2+4k+1=8N+1$,
we deduce that $(2k+1)^4=(8N+1)^2=64N^2+16N+1=16M+1$ implies that $(2k+1)^4-1=0$ mod $16$.
I decided to try and do it by binomial expansion as op suggested...
Let $n = 2k+1$ and compute: \begin{align} n^5 -n &= (2k+1)((2k+1)^4 - 1)\\ &= (16k^4 + 32k^3 + 24k^2 + 8k)(2k+1) \\ &= 8k(2k^3 + 4k^2 + 3k + 1)(2k+1) \\ &= 8k(k+1)(1+2k+2k^2)(2k+1)\\ &= 8\ k(k+1)\ \big(5 + 2(k-1)(k+2)\big)\ \big(5+2(k-2)\big). \end{align} $k(k+1)$ is even and exactly one of $k-2,k-1,k,k+1,k+2$ is divisible by 5 so $n^5-n$ is divisible $8\times2=16$ and 5.
$\hspace{18cm}\square$
Definitely don't recommend this strategy :-P
