For every $n\in \mathbb N$ we have
\begin{equation*}
\lfloor\sqrt{4n+1} \rfloor = \lfloor\sqrt{4n+2} \rfloor.
\end{equation*}
The proof is by cases. Assume first that $m = \lfloor\sqrt{4n+1} \rfloor$ is even. Then $(m+1)^2$ is an odd square and $(m+1)^2 - (4n+1) \gt 0$. Since $(m+1)^2$ leaves remainder $1$ when divided by $4$ we deduce that $(m+1)^2 - (4n+1) \gt 3$. Therefore, $(m+1)^2 - (4n+2) \gt 2$. Consequently,
$$
m^2 \leq 4n+1 \lt 4n+2 \lt (m+1)^2
$$
and therefore $\lfloor\sqrt{4n+2}\rfloor = m$. Now assume that $m = \lfloor\sqrt{4n+1} \rfloor$ is odd. Then $(m+1)^2$ is an even square and $(m+1)^2 - (4n+1) \gt 0$. Since $(m+1)^2$ is divisable by $4$ we deduce that $(m+1)^2 - (4n+1) \gt 2$. Therefore, $(m+1)^2 - (4n+2) \gt 1$. Thus again,
$$
m^2 \leq 4n+1 \lt 4n+2 \lt (m+1)^2
$$
and therefore $\lfloor\sqrt{4n+2}\rfloor = m$. This proves the stated equality.
This inequaltiy, together with the inequality
$$
\forall\, n \in{\mathbb N} \quad \sqrt{4n+1} \lt \sqrt{n}+\sqrt{n+1} \lt \sqrt{4n+2}
$$
which has been proved in earlier answers, implies
$$
\forall\, n \in{\mathbb N} \quad \lfloor\sqrt{n}+\sqrt{n+1}\rfloor = \lfloor\sqrt{4n+1}\rfloor.
$$