1. ## floor equality

Prove that :

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

2. 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} (meaning that their integer part is the same)

Square it :
$2n+1+2\sqrt{n(n+1)}
$2n+1+2 \sqrt{n^2+n}

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

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.

3. hello
thanks Moo

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

4. 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 --> nČ=4kČ, which is 0 modulo 4
- if it's odd : n=2k+1 --> nČ=4kČ+4k+1, which is 1 modulo 4