<s>[INST] Mathematical Induction Involving The Floor Function I need to prove the identity $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$ for all natural numbers $n$.  
I wanted to use mathematical induction. The identity is true for $n=1$. Then, I assume $$\lfloor \sqrt{k}+\sqrt{k+1}\rfloor=\lfloor\sqrt{4k+2}\rfloor$$
and need to show that it is also true for $n=k+1$
$$\lfloor \sqrt{k+1}+\sqrt{k+2}\rfloor=\lfloor\sqrt{4k+6}\rfloor.$$
I thought about the inequality $m\leq\sqrt{4k+6}<m+1$ for some integer $m$, and the same for the LHS, but this doesn't seem to help and I don't have other ideas.
 [/INST] Here is an (maybe) easier approach.
To prove $\lfloor\sqrt n+\sqrt {n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$, and it's easy to notice that $\lfloor\sqrt{4n+2}\rfloor>\lfloor\sqrt n+\sqrt {n+1}\rfloor$ and that $\lfloor\sqrt{4n+2}\rfloor-\lfloor\sqrt n+\sqrt {n+1}\rfloor<1$, it suffices to prove that there's no integer between  $\sqrt n+\sqrt {n+1}$ and $\sqrt{4n+2}$.

We suppose that there is an integer $k$ between $\sqrt n+\sqrt {n+1}$ and $\sqrt{4n+2}$. Then
\begin{equation*}
\sqrt n+\sqrt {n+1}\leq k\leq\sqrt{4n+2}\\
n+n+1+2\sqrt{n(n+1)}\leq k^2\leq 4n+2
\end{equation*}
Since $\sqrt{n(n+1)}>n$, we have
\begin{equation*}
4n+1<k^2\leq4n+2
\end{equation*}
So the only possibility is $k^2=4n+2$. So $k=\sqrt{4n+2}=2\sqrt{n+1/2}=2\sqrt{(2n+1)/2}$. So
\begin{equation}
\sqrt{\frac{2n+1}{2}}=x\quad (1),\quad or\\
\sqrt{\frac{2n+1}{2}}=y+\frac{1}{2}\quad(2)
\end{equation}
for integers $x$ and $y$. Since $2n+1$ is an odd number, $(2n+1)/2$ cannot be an integer, so $(1)$ cannot be the case. For $(2)$, we notice that $(y+1/2)^2=(2y+1)^2/4$, which is an odd number over $4$. But $(2n+1)/2$ is an odd number over $2$, namely an even number over $4$, thus they can not be equal. We have the desired contradiction.
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