Why does the difference between squares of the $i^{th}$ even and odd integers increase by 4? I've been trying to find a formula for sum of the first n squares of odd numbers (a question from Spivak's calculus) and I was trying to subtract the the $i^{th}$ odd number from the $i^{th}$ even number, I've noticed that the the difference between the squares is (or seems to be) increasing by 4. This what I'm seeing:
\begin{align*}
2 ^ 2 - 1 ^ 2 = 3\\
4 ^ 2 - 3 ^ 2 = 7 \\
6 ^ 2 - 5 ^ 2 = 11 \\
8 ^ 2 - 7 ^ 2 = 15 \\
10 ^ 2 - 9 ^ 2 = 19 \\
12 ^ 2 - 11 ^ 2 = 23
\end{align*}
Every consecutive term in $3,7,11,15,19,23$ differs by 4. The fact that the difference of the squares for the $i^{th}$ even and odd integers increases by 4 seems completely arbitrary. Why 4? Is there something going on that I'm just missing? I don't know if it has something to do with Dirichlet's arithmetic progression for $4n + 3$. The pattern indicates this formula to me:
$$ \sum_{k=1}^{n} (2k)^2 = \sum_{k=1}^{n}(2k-1)^2 + \sum_{k=0}^{n-1}(4k+3) $$
I believe this formula is true (though I haven't proved it yet), this will make it possible for me to use the closed form: $$\sum_{k=1}^{2n}k^2 = \frac{n(2n+1)(4n+1)}{3}$$
to find the closed form for the sum of odd numbers, But the question is: why is the difference between the $n^{th}$ even and odd integer equal to $4(n-1)+3$? 

Choose an even number and write as $2k$. Then the first two differences between 
even/odd squares are:
$4k^2 - (2k - 1)^2 = 4k^2 - ( 4k^2 - 4k +1) = 4k - 1$ 
$ (2k+2)^2 - (2k+1)^2 = 4k^2 +8k + 4 - (4k^2 +4k  + 1) = 4k + 3 $
and clearly $4k+3 - (4k-1) = 4$
