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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.

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OP: You might want to stop systematically defacing your questions, the fact has been signalled to the moderators and may result in your suspension. –  Did May 6 '13 at 19:43

2 Answers 2

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.

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Take $n=2k+1$ then,

$\lfloor(n^2/4)\rfloor=\lfloor k^2+k+1/4\rfloor=k^2+k$

$\frac{(n-1)(n+1)}{4}=(n^2-1)/4=k^2+k=\lfloor(n^2/4)\rfloor$

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