proving for all odd integers that $n^2 + 2n \equiv 0 \pmod{3}$

Question

prove that for all odd integers, $3 |(n^2 + 2n)$

An even integer may be described as $2k$ and an odd one as $(2k+1)$, inserting it in to our equation gives us

$(2k+1)^2 + 2(2k+1) $

$=4k^2 + 8k + 3$

$=2(2k^2 + 4k) + 3$

what is the next step to prove this?

Answer 1

For $n=5$ you have $$n^2+2n=25+10 = 5 \times 7 = 35 \equiv 2 \pmod 3.$$

In general for $n= 6k-1$ you have $$(6k-1)^2+2(6k-1) = (6k-1)(6k+1) = 36k^2-1 \equiv 2 \pmod 3.$$

Answer 2

We need $3|(n^2+2n)=n(n+2)$

As $3$ will exactly divide one of $n, n+1, n+2$ (why?)

$\implies3|n$ or $3|(n+2)$

So if $3|(n+1),\iff3\nmid n(n+2)$

Answer 3

$(2n+1)^2+2(2n+1)=4n^2+8n+3\equiv n^2+2n=n(n+2) \pmod 3$.

You can choose $n$ such that neither $n$ nor $n+2$ be divisible by $3$. Your proposition is not true.