Prove that if $4k+1$ is a prime number than for every $n \in \mathbb{N}$ $\sum\limits_{j=1}^{2k}{j^{4n+2}}$ is a multiple of $4k+1$
First I tried proving for n=1
$\sum\limits_{j=1}^{2k}{j^{6}}=(4k+1)a$ ... (1)
So I realized that
$7\sum\limits_{j=1}^{k}{j^{6}}=(k+1)^7-21\sum\limits_{j=1}^{k}{j^{5}}-35\sum\limits_{j=1}^{k}{j^{4}}-35\sum\limits_{j=1}^{k}{j^{3}}-21\sum\limits_{j=1}^{k}{j^{2}}-7\sum\limits_{j=1}^{k}{j}$
After some calculations I arrived to the conclusion that
$\sum\limits_{j=1}^{k}{j^{6}}= \frac{k(2k+1)(4k+1)(48k^4+48k^3-6k+1)}{21}$
Now we can prove for n=1
On LHS of (1) we have
$\frac{k(2k+1)(4k+1)(48k^4+48k^3-6k+1)}{21}+\frac{k(2k+1)(4k+1)(48k^4+48k^3-6k+1)}{21}$
we can just factor out $4k+1$ In this sum so therefore the statement for n=1 is true
now we must assume that the statement is true for n=a $a\in \mathbb{N}$
$$\sum\limits_{j=1}^{2k}{j^{4a+2}}=(4k+1)b$$
now we must prove the statement is true for $n=a+1$
I've tried some things but ultimately I can't get anywhere after this point. Any help is appreciated