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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.

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Let

$$m = \sqrt{n} + \sqrt{n + 1} \tag{1}\label{eq1A}$$

Since $m$ is a positive quantity, you have that $m = \sqrt{m^2}$, so you get

$$\begin{equation}\begin{aligned} m & = \sqrt{\left(\sqrt{n} + \sqrt{n + 1}\right)^2} \\ & = \sqrt{n + 2\sqrt{n(n+1)} + n + 1} \\ & = \sqrt{2n + 1 + 2\sqrt{n(n+1)}} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Since you also have for all natural numbers $n$ (i.e., $n \gt 0$) that

$$n \lt \sqrt{n(n+1)} \lt n + 1 \tag{3}\label{eq3A}$$

Thus, you also have from \eqref{eq2A} that

$$\begin{equation}\begin{aligned} \sqrt{2n + 1 + 2(n)} & \lt \sqrt{2n + 1 + 2\sqrt{n(n+1)}} \lt \sqrt{2n + 1 + 2(n + 1)} \\ \sqrt{4n + 1} & \lt m \lt \sqrt{4n + 3} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

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 these values must be less than or equal to $4n + 1$, say it's $k^2$. Thus, you have that

$$k \le \sqrt{4n + 1} \lt m \lt \sqrt{4n + 3} \lt k + 1 \tag{5}\label{eq5A}$$

In summary, you thus have that

$$\lfloor m \rfloor = \lfloor \sqrt{n} + \sqrt{n + 1} \rfloor = \lfloor \sqrt{4n + 2} \rfloor = k \tag{6}\label{eq6A}$$

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