<s>[INST] Proving an identity in floor function For every positive integer $n$ ,show that 
$$\lfloor{\sqrt{n}+\sqrt{n+1}}\rfloor=\lfloor{\sqrt{4n+1}}\rfloor=\lfloor{\sqrt{4n+2}}\rfloor=\lfloor{\sqrt{4n+3}}\rfloor$$
My Attempt:
$$\left(\sqrt{n}+\sqrt{n+1}\right)^2=2n+1+2\sqrt{n(n+1)}$$
Now, $\sqrt{n(n+1)}$ is geometric mean of $n$ and $n+1$, therefore
$$n<\sqrt{n(n+1)}<n+1$$
$$2n<2\sqrt{n(n+1)}<2(n+1)$$
$$4n+1<2n+1+2\sqrt{n(n+1)}<4n+3$$
$$4n+1<\left(\sqrt{n}+\sqrt{n+1}\right)^2<4n+3$$
But after this not able to get anywhere
 [/INST] Note that for an integer, $x$, $$x^2 \equiv 0 \pmod 4$$ or
$$x^2 \equiv 1 \pmod 4$$
Hence there is no complete square between $4n+1$ and $4n+3$.
Hence $\lfloor \sqrt{4n+1} \rfloor = \lfloor \sqrt{4n+3} \rfloor$, the overall task can be completed by your earlier result of $4n+1 < (\sqrt{n} + \sqrt{n+1})^2 < 4n+3$.
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