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.
 A: The identity is true for $n = 0$, so consider $n \ge 1$. Also, let
$$m = \sqrt{n} + \sqrt{n + 1} \tag{1}\label{eq1A}$$
With $m \gt 0 \; \to \; m = \sqrt{m^2}$, we get
$$\begin{equation}\begin{aligned}
m & = \sqrt{\left(\sqrt{n} + \sqrt{n + 1}\right)^2} \\
& = \sqrt{n + 2\sqrt{n(n+1)} + n + 1} \\
& = \sqrt{2n + 1 + 2\sqrt{n(n+1)}}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
We also have
$$\color{blue}{n} \lt \color{red}{\sqrt{n(n+1)}} \lt \color{green}{n + 1} \tag{3}\label{eq3A}$$
Thus, using this with \eqref{eq2A} results in
$$\begin{equation}\begin{aligned}
\sqrt{2n + 1 + 2(\color{blue}{n})} & \lt \sqrt{2n + 1 + 2\color{red}{\sqrt{n(n+1)}}} \lt \sqrt{2n + 1 + 2(\color{green}{n + 1})} \\
\sqrt{4n + 1} & \lt m \lt \sqrt{4n + 3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since natural numbers squared are congruent to $0$ modulo $4$ for even values and to $1$ modulo $4$ for odd values, neither $4n + 2$ or $4n + 3$ can be a perfect square. Thus, the largest perfect square less than or equal to values in \eqref{eq4A}, say it's $k^2$, must be less than or equal to $4n + 1$. This therefore gives
$$k \le \sqrt{4n + 1} \lt m \lt \sqrt{4n + 3} \lt k + 1 \tag{5}\label{eq5A}$$
Finally, we get that
$$k = \lfloor m \rfloor = \lfloor \sqrt{n} + \sqrt{n + 1} \rfloor = \lfloor \sqrt{4n + 2} \rfloor \tag{6}\label{eq6A}$$
A: 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.
