Prove that if an integer $n$ is odd, then there exists some integer $c$ such that $n^2 = 4c + 1$

Question

So far, since we know that $n$ is an odd integer, we can set $n = 2k + 1$ by definition of odd integers.

Now we have $$(2k + 1)^2 = 4c + 1 \implies k^2 + k = c$$

and I am currently stuck on the step highlighted above.

Answer 1

You're nearly done. As you said, if $n$ is odd, then $n=2k+1$ for some integer $k$. So $$n^2=4k^2+4k+1=4(k^2+k)+1=4c+1$$ So $c=k^2+k$ satisfies the equation. So all you should do is just eliminate the $4$ when you are simplifying.

Answer 2

More is true:

If $n$ is odd there exists $c$ such that $n^2 = 8c+1$.

Proof:

Since $n$ is odd, $n = 2m+1$ so that $n^2 = (2m+1)^2 =4m^2+4m+1 =4m(m+1)+1 $.

Whether $m$ is even or odd, $m+1$ has opposite parity, so $m(m+1)$ is even.

Therefore, if we let $c = \dfrac{m(m+1)}{2}$, then $c$ is an integer and $n^2 = 8c+1$.