TITLE: Show that $n^2+n$ is even for all $n\in\mathbb{N}$ by contradiction
QUESTION [5 upvotes]: Show that $n^2+n$ is even for all $n\in\mathbb{N}$ by contradiction.

My attempt: assume that $n^2+n$ is odd, then $n^2+n=2k+1$ for all $k\in\mathbb{N}$.
We have that 
$$n^2+n=2k+1 \Leftrightarrow \left(n+\frac{1}{2}\right)^2-\frac{1}{4}=2k+1\Leftrightarrow n=\sqrt{2k+\frac{5}{4}}-\frac{1}{2}.$$
Choosing $k=1$, we have that $n=\sqrt{2+\frac{5}{4}}-\frac{1}{2}\notin\mathbb{N}$, so we have a contradiction because $n^2+n=2k+1$ should be verified for all $n\in\mathbb{N}$ and for all $k\in\mathbb{N}$.
Is this correct or wrong? If wrong, can you tell me where and why? Thanks.

REPLY [2 votes]: To be different.  Suppose there some natural number where this is true.  But it isn't true for $1$ or $2$.  But if there is some value where it is true there must be some first value, $n$ where it is true.
So suppose that $n^2 + n$ is odd but $(n-1)^2 + (n-1)$ is not odd.  Then $n^2 - 2n + 1+ n-1 = n^2-n + 2$ is even. So $n^2 - n + 2 = 2k$ for some $k$.
So $n^2 - n + 2 + (2n-2) = 2k + (2n-2) = 2(k+n -1)$ is even.  But $n^2 - n + 2 + (2n-2)=n^2 + n$ which we assumed was odd.
The advantage of this proof is that we don't have to assume that if $n^2 + n = n(n+1)$ is odd that means $n$ and $n+1$ are both odd (why is that so?) and that it is impossible for $n$ and $n+1$ to both be odd (why is that so?).
The disadvantage of this proof is it's abstract and not straightforward.