floor equality

**linda2005** · April 23rd 2009, 10:11 AM

Hello, Thanks for your help
Prove that :

$\forall n\in\mathbb{N}^{*},\;\lfloor\sqrt{n}+\sqrt{n+1}\rf loor =\lfloor\sqrt{4n+2}\rfloor.$

**Moo** · April 24th 2009, 12:30 AM

Hello,

I found it somewhere

First prove that $\sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$

$\begin{aligned} 4n^2+4n<4n^2+4n+1 & \Leftrightarrow (2\sqrt{n(n+1)})^2<(2n+1)^2 \\ & \Leftrightarrow 2\sqrt{n(n+1)}<2n+1 \\ & \Leftrightarrow n+(n+1)+2 \sqrt{n} \sqrt{n+1}<4n+2 \\ & \Leftrightarrow (\sqrt{n}+\sqrt{n+1})^2<4n+2 \\ & \Leftrightarrow \sqrt{n}+\sqrt{n+1}<\sqrt{4n+2} \end{aligned}$

Then prove that there exists no integer k such that $\sqrt{n}+\sqrt{n+1}<k\leq \sqrt{4n+2}$ (meaning that their integer part is the same)

Square it :
$2n+1+2\sqrt{n(n+1)}<k^2 \leq 4n+2$
$2n+1+2 \sqrt{n^2+n}<k^2\leq 4n+2$

From $\sqrt{n^2+n}<\sqrt{n^2}=n$ , we can say that $2n+2n+1=4n+1<2n+1+2 \sqrt{n^2+n}$

Thus we finally have :
$4n+1<k^2\leq 4n+2$

Since k is an integer, $k^2$ is also an integer.
Thus $k^2=4n+2$
But a squared number can only be 0 or 1 modulo 4.
Therefore, this integer k doesn't exist.

**linda2005** · April 24th 2009, 12:52 PM

hello
thanks Moo

why the squared number can only be 0 or 1 modulo 4 ? you can give me the proof

**Moo** · April 25th 2009, 12:20 AM

Originally Posted by linda2005

hello
thanks Moo

why the squared number can only be 0 or 1 modulo 4 ? you can give me the proof

Here is, imho, the simplest way :
a number can only be even or odd.
- if it's even : n=2k --> n²=4k², which is 0 modulo 4
- if it's odd : n=2k+1 --> n²=4k²+4k+1, which is 1 modulo 4

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