<s>[INST] Proof Regarding Property of Odd Integers The question I am working on is:

"Use a direct proof to show that every odd integer is the difference of two squares."

Proof:
Let n be an odd integer: $n = 2k + 1$, where $k \in Z$
Let the difference of two different squares be, $a^2-b^2$, where $a,b \in Z$.
Hence, $n=2k+1=a^2-b^2$...
As you can see, this a dead-end. Appealing to the answer key, I found that they let the difference of two different squares be, $(k+1)^2-k^2$. I understand their use of $k$--$k$ is one number, and $k+1$ is a different number--;however, why did they choose to add $1$? Why couldn't we have added $2$?
 [/INST] Another approach, a graphical proof:
$$\underbrace{\begin{array}{ccccc}
1\odot & 3\otimes & 5\odot & \cdots & (2k - 1)\otimes\\\hline
\bigodot & \bigotimes & \bigodot & \cdots & \bigotimes\\
\bigotimes & \bigotimes & \bigodot & \cdots & \bigotimes\\
\bigodot & \bigodot & \bigodot & \cdots & \bigotimes\\
\vdots & \vdots & \vdots & \ddots & \vdots\\[1ex]
\bigotimes & \bigotimes & \bigotimes & \cdots & \bigotimes\\
\end{array}}_{\textstyle k}$$
From this we have $k^2 = 1 + 3 + 5 + \dotsb + (2k - 1)$. Hence,
$$\begin{align}
2k + 1 &= \Big(1 + 3 + 5 + \dotsb + (2k - 1) + (2k + 1)\Big) - \Big(1 + 3 + 5 + \dotsb + (2k - 1)\Big)\\
       &= (k + 1)^2 - k^2
\end{align}$$
</s>