Proving that if $n$ is odd, then $n^2=8k+1 \text{ for some } k\in\mathbb{Z}$

Question

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?

Answer 1

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 $$ n^2 = 4m^2+4m+1 = 4m(m+1)+1 = 4(2k)+1 = 8k+1 $$

Answer 2

hint: Consider mod $4$, $n = 4k+1$ since it's an odd integer.

Answer 3

You can just say that, within two consecutive numbers, there'll always be one even. Thus, $m\cdot(m+1)$ is an even number that can be written as $2k$, $k\in\mathbb{Z}$.

Answer 4

Your proof is correct.

You may consider $n=4m\pm1$. The proof will be a bit shorter.

Answer 5

just consider the 2 cases for m. m can be even or m can be odd. so for case 1 let m=2x and plug that into the equation and you get an even number. case 2 let m=2x+1 and plug into the equation and you get an even number again. so for both cases the number is even and therefore proven.