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$\begingroup$

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.

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    I don't know what step is highlighted. But what about expanding $(2k+1)^2$ and defining $c$ such that it makes sense? Also, I don't understand what you mean by the equal sign in the parentheses. – 2017-01-27
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    Let $n=2k+1$. Then $n^2 = (2k+1)^2 = 4k^2+4k+1 = 4(\underbrace{\dots}_{\text{call this }~c})+1$ – 2017-01-27
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    note that $n^2 = 4c + 1$ is equivalent to $n^2 - 1^2 = 4c$ – 2017-01-27

2 Answers 2

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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.

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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 $n^2 = 8c+1$.

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    You realize that you've just said 2=1, right? If n^2 equals both 4c+1 and 8c+1, then we can set them equal to each other, subtract one from both sides, and divide both sides by 4c to get 2=1. So obviously something is wrong: how can n^2 equal both? – 2017-01-27
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    @DonielF: You realize that $c$ in this proof is not the same as $c$ in the original statement? – 2017-01-27
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    Yup. It's half. – 2017-01-27