Proving that if $8\mid (n^2+2n)$ then $2\mid n$

Question

Let $n\in \mathbb N$ prove that if $8\mid (n^2+2n)$ then $2\mid n$.

From the given, there exists $k\in \mathbb N$ such that $8k= (n^2+2n)$, take $k=1$, and we get $2\cdot 4 = n(n+2)$.

Now my question is, can I do the following?

$2\cdot 4 = n(n+2)\to 2\mid n(n+2)\to (2\mid n)\wedge (2\mid n+2)$ thus $2\mid n$

Answer 1

Hints:

$$n^2+2n=n(n+2)$$

But $\;n\,,\,\,n+2\;$ have the same parity and an even number divides the above, so...

Answer 2

if $n=2k+1$ then $n^2+2n$ is odd number..

if $n=2k$ then $n^2+2n=(2k)^2+2.(2k)=4k^2+4k=4k(k+1)$, but $k(k+1)$ is even number so that, we have done.

Answer 3

Hint:

If $n$ is odd then so is $n^2$ and also of course $n^2+2n$.

Answer 4

$$n^2+2n\equiv0\mod{8}\Rightarrow n^2+2n\equiv0\mod{4}\Rightarrow (n+1)^2-1\equiv0\mod{4}$$

$$\Rightarrow (n+1)^2\equiv1\mod{4}\Rightarrow n+1\equiv1\mod{2}\Rightarrow n\equiv0\mod{2}$$