TITLE: How to prove that $k^3+3k^2+2k$ is always divisible by $3$?
QUESTION [13 upvotes]: How can I prove that the following polynomial expression is divisible by 3 for all integers $k$?
$$k^3 + 3k^2 + 2k$$

REPLY [2 votes]: By Fermat's Little theorem $k^3\equiv k\pmod{3}$, therefore:
$$k^3+\underbrace{3k^2}_{\equiv 0}+\underbrace{2k}_{\equiv -k}\equiv k^3-k\equiv 0\pmod{3}$$