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 $2$?
 A: 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,
$$\begin{align}
2k + 1 &= \Big(1 + 3 + 5 + \dotsb + (2k - 1) + (2k + 1)\Big) - \Big(1 + 3 + 5 + \dotsb + (2k - 1)\Big)\\
       &= (k + 1)^2 - k^2
\end{align}$$
A: Directly:
$$2k+1=k^2+2k+1-k^2=(k+1)^2-k^2\ldots$$
A: Here you have another approach...
Note that
$$
  \sum_{i=1}^n
  {
    i
  }
=
  \frac{n(n+1)}2
$$
so that the sum of the first $n$ odd naturals is
$$
  \sum_{i=1}^n
  {
    (2i-1)
  }
=
  2\left(
    \sum_{i=1}^n
    {
      i
    }
  \right)
- n
=
  n(n+1)-n
=
  n^2
$$
We have showed that the sum of the first $n$ odd naturals is $n^2$. The $n$-th odd natural is, trivially, equal to the sum of the first $n$ minus the sum of the first $n-1$:
$$
  2n-1
=
  \sum_{i=1}^n
  {
    (2i-1)
  }
  -
  \sum_{i=1}^{n-1}
  {
    (2i-1)
  }
$$
and from what we found, we can write it as
$$
2n-1 = n^2 - (n-1)^2
$$
A: What's going on here, I think, is you're confused about what you're being asked to prove.
The statement is "every odd integer is the difference of two squares", or, more precisely, 
 "for all odd integers $n$, there exist $a$ and $b$ such that $a^2 - b^2 = n$". Think for a bit about what the difference between "for all..." and "there exists..." is, and maybe you'll realise what's going on.
You have to come up with a proof that works for every odd integer, so you start with an arbitrary choice $n = 2k + 1$. But then you seem to go on to pick an arbitrary difference of two squares, $a^2 - b^2$. But that's not necessary – we're not trying to prove anything about every difference of two squares, just one, so you only need to show that there is a difference of two squares that satisfies the condition, and in particular you choose it, it isn't arbitrary.
Think of the statement as "if you give me an odd integer $n$, I can give you $a$ and $b$ such that $a^2 - b^2 = n$". In particular, you can give me whatever odd integer you like, but then I get to choose what $a$ and $b$ are. In particular, I can choose $a$ to be $b + 1$ if I like.
From one of your comments:

We have to show that $a^2 - b^2$ is an odd number

But in general that's just not true. For example, $4^2 - 2^2 = 12$. So if you've found yourself trying to prove that statement, you've clearly gone wrong somewhere.
A: Note that $a^2 - b^2 = (a + b)(a - b)$. Solve the simultaneous equations $a + b = n$ and $a - b = 1$.
This is where you got $(k + 1)^2 - k^2$ from - $(a + b)(a - b)$ then matches to $((k + 1) + k)((k + 1) - k)$.
A: $$2k + 1 = 1 \times (2k  +1) = (\color{#C00}{k + 1} -{\bf k}) (\color{#C00}{k + 1} +\bf{ k})$$
