First, i will show, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+3} \right\rfloor$, what is equivalent to that: $ \sqrt{4n+3} - \sqrt{4n+1} < 1 \\ 4n+3+4n+1-2\sqrt{(4n+1)(4n+3)} < 1 \\ 4n+2-\sqrt{(4n+1)(4n+3)}< 1/2$ (Of course, this difference is greater than $0$.) It is obvious, that $4n+5/3 < \sqrt{(4n+1)(4n+3)}$ for all $n>0$, so you can say: $4n+2-(4n+5/3)<1/2$, what is true. So it is shown, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+3} \right\rfloor$. Now, because of $\sqrt{4n+1} \leq \sqrt{4n+2} \leq \sqrt{4n+3}$ and $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+3} \right\rfloor$, it's also true, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+2} \right\rfloor =\left\lfloor \sqrt{4n+3} \right\rfloor$ For $ \sqrt{n}+\sqrt{n+1} $ holds the same condition as for $\sqrt{4n+2}$, i.e. $ \sqrt{4n+1} \leq \sqrt{n}+\sqrt{n+1} \leq \sqrt{4n+3} $, so that you can also say, that $\left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{n}+\sqrt{n+1} \right\rfloor =\left\lfloor \sqrt{4n+3} \right\rfloor$, so the proposition is proven.

To prove that $ \left\lfloor \sqrt{4n+1} \right\rfloor = \left\lfloor \sqrt{4n+3} \right\rfloor$, actually we only need to show that there is no integer in $ (\sqrt{4n+1},\sqrt{4n+3}]$, i.e., there is no perfect square in $ (4n+1,4n+3]$, which is quite obvious.

Hello, This is kinda old but I liked this problem and I will give a very beautiful solution for it: $\begin{array}{l} n = \sqrt {n^2 } < \sqrt {n\left( {n + 1} \right)} < \sqrt {\left( {n + 1} \right)^2 } = n + 1 \\ \\ \Rightarrow 2n < 2\sqrt {n\left( {n + 1} \right)} < 2n + 2 \to (1) \\ \\ 2\sqrt {n\left( {n + 1} \right)} = \left( {\sqrt n + \sqrt {n + 1} } \right)^2 - \left( {2n + 1} \right) \to (2) \\ \\ (2)in(1) \Rightarrow 2n < \left( {\sqrt n + \sqrt {n + 1} } \right)^2 - \left( {2n + 1} \right) < 2n + 2 \\ \\ \Rightarrow 4n + 1 < \left( {\sqrt n + \sqrt {n + 1} } \right)^2 < 4n + 3 \\ \\ \Rightarrow \sqrt {4n + 1} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 3} \to (3) \\ \\ \exists k \in Z^ + :k^2 \le 4n + 1 < \left( {k + 1} \right)^2 \\ \\ But \\ \\ 4n + 2 \wedge 4n + 3\left( { \sim \equiv } \right)2,3\left( {\bmod 4} \right) \Rightarrow 4n + 2 \wedge 4n + 3 \ne \left( {k + 1} \right)^2 \\ \\ \Rightarrow k^2 \le 4n + 1 < 4n + 2 < 4n + 3 < \left( {k + 1} \right)^2 \\ \\ \Rightarrow k \le \sqrt {4n + 1} < \sqrt {4n + 2} < \sqrt {4n + 3} < k + 1 \\ \\ (3) \Rightarrow \sqrt {4n + 1} ,\sqrt {4n + 2} ,\sqrt {4n + 3} ,\sqrt n + \sqrt {n + 1} \in \left[ {k,k + 1} \right) \\ \\ \Rightarrow \left\lfloor {\sqrt n + \sqrt {n + 1} } \right\rfloor = \left\lfloor {\sqrt {4n + 1} } \right\rfloor = \left\lfloor {\sqrt {4n + 2} } \right\rfloor = \left\lfloor {\sqrt {4n + 3} } \right\rfloor = k \\ \end{array}$