Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by 4 I'm studding D.M Burton & want to solve:  Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by $4$. . Please help me by giving your solution to it. I'm new comer to number theory so please don't use theorems above Theory of Congruence.

Another way is to use the "paired element" approach...  Since $a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$, we have
$$1^5+99^5+2^5+98^5+\dots+ 50^5+100^5\\
=100(1-99+99^2-99^3+99^4)+100(2^4-\dots)+\dots+100(49^4-\dots)+2^5\cdot(25)^5+100^5\\
\equiv 0\pmod 4$$
