Proving that for any odd integer:$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$ I'm trying to figure out how to prove that for any odd integer, the floor of:
$$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$$  
Any help is appreciated to construct this proof!  
Thanks guys.

Let $n$ be an odd integer.
Then there exists an integer $k$, such that:
$$n=2k+1$$
It follows that:
$$\begin{align}
\left\lfloor\frac{n^2}{4}\right\rfloor &= \left\lfloor\frac{(2k+1)^2}{4}\right\rfloor\\
&=\left\lfloor\frac{(4k^2+4k+1)}{4}\right\rfloor\\
&=\left\lfloor\frac{(4k^2+4k)}{4}+\frac{1}{4}\right\rfloor\\
&=\left\lfloor(k^2+k)+\frac{1}{4}\right\rfloor
\end{align}
$$
Because $k^2+k$ is an integer, we can now say:
$$\left\lfloor\frac{n^2}{4}\right\rfloor = k^2+k$$
It also follows that:
$$\begin{align}
\frac{(n-1)(n+1)}{4} &= \frac{n^2-1}{4}\\
&= \frac{(2n+1)^2-1}{4}\\
&= \frac{(4k^2+4k+1)-1}{4}\\
&= \frac{4k^2+4k}{4}\\
&= k^2+k\\
\end{align}$$
Therefore:
$$\left\lfloor\frac{n^2}{4}\right\rfloor=\frac{(n-1)(n+1)}{4}$$
Q.E.D.
