Why does this pattern occur? I first saw this pattern when I was trying to factor quadratics.
let $a + b = k$ where $k$ is any constant,
Now let $(a - b)/2 = x$
It appears to be that   $(k^2/4) - x^2 = a*b$
For example, let's say $k = 8$
then,
$4*4 = 16$
$3*5 = 15$
$2*6 = 12$
$1*7 = 7$
$0*8 = 0$
$-1*9 = -9$
If we subtract 16 from all of those, we get $0, -1, -4,-9, -16, and -25$.
My question is, why does this pattern occur? Maybe I'm missing something obvious, but there are so many variables that I don't know where to start

In the original post, $x=a-b$ and was changed to $x=\frac{a-b}{2}$.  This answer addresses both formulae.
Consider
$$
\frac{k^2}{4}-x^2=\frac{(a+b)^2}{4}-(a-b)^2=\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}-a^2+2ab-b^2=-\frac{3}{4}a^2+\frac{5}{2}ab-\frac{3}{4}b^2=-\frac{3}{4}(a-b)^2+ab
$$
This is not $ab$, except in fairly special situations.
Now, if $x=\frac{a+b}{2}$ (as suggested in the comments and what appears to be the OP's original intent from the line starting with if we subtract 16...)
$$
\frac{k^2}{4}-x^2=\frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}=\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}-\frac{a^2}{4}+\frac{ab}{2}-\frac{b^2}{4}=ab,
$$
as desired.
This happens because $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$.  Therefore, $(a+b)^2-(a-b)^2=4ab$, which can be scaled to $ab$.  Tricks like this often show up on calculus exams, especially in arc-length situations.
