I'm currently studying Lay's Analysis with an introduction to Proofs.
I am trying to prove this statement:
If $n$ is odd, then there exists a $k\in\mathbb{Z}$ such that $n^2=8k+1$
I first started by saying that $n$ is odd, then it can be written as $n=2m+1$ for $m\in \mathbb{N}$. Substituting this in, I find that
$$4m^2+4m+1=8k+1$$
$$m^2+m=2k$$
$$m(m+1)=2k$$
I now just need to show that $m(m+1)\text{ for } m\in\mathbb{N}$ is even, which I am not sure how to rigorously prove.
Is it enough to say: if $m$ is odd, then $m+1$ is even, thus their product is even; if $m$ is even, then $m+1$ is odd, and their product is once again even? Their product is even because multiplying a number that is divisible by $2$ with any integer yields a product that $2$ still divides.
How can I improve my argument?
Best Answer
As I commented, what you are doing is sufficient. Since you are new to proofs, I figured I would write out the proof in a logical order.
Since $n$ is odd, there exists a non-negative integer $m$ such that $n = 2m+1$. Furthermore, (by your argument), there exists another non-negative integer $k$ such that $m(m+1) = 2k$. Now, we may write $$ n^2 = 4m^2+4m+1 = 4m(m+1)+1 = 4(2k)+1 = 8k+1 $$