Hello, Thanks for your help
Prove that :
$\displaystyle \forall n\in\mathbb{N}^{*},\;\lfloor\sqrt{n}+\sqrt{n+1}\rf loor =\lfloor\sqrt{4n+2}\rfloor.$
Hello,
I found it somewhere
First prove that $\displaystyle \sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}$
$\displaystyle \begin{aligned}
4n^2+4n<4n^2+4n+1 & \Leftrightarrow (2\sqrt{n(n+1)})^2<(2n+1)^2 \\
& \Leftrightarrow 2\sqrt{n(n+1)}<2n+1 \\
& \Leftrightarrow n+(n+1)+2 \sqrt{n} \sqrt{n+1}<4n+2 \\
& \Leftrightarrow (\sqrt{n}+\sqrt{n+1})^2<4n+2 \\
& \Leftrightarrow \sqrt{n}+\sqrt{n+1}<\sqrt{4n+2}
\end{aligned}$
Then prove that there exists no integer k such that $\displaystyle \sqrt{n}+\sqrt{n+1}<k\leq \sqrt{4n+2}$ (meaning that their integer part is the same)
Square it :
$\displaystyle 2n+1+2\sqrt{n(n+1)}<k^2 \leq 4n+2$
$\displaystyle 2n+1+2 \sqrt{n^2+n}<k^2\leq 4n+2$
From $\displaystyle \sqrt{n^2+n}<\sqrt{n^2}=n$, we can say that $\displaystyle 2n+2n+1=4n+1<2n+1+2 \sqrt{n^2+n}$
Thus we finally have :
$\displaystyle 4n+1<k^2\leq 4n+2$
Since k is an integer, $\displaystyle k^2$ is also an integer.
Thus $\displaystyle k^2=4n+2$
But a squared number can only be 0 or 1 modulo 4.
Therefore, this integer k doesn't exist.