Proving an identity in floor function

Question

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

Answer 1

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 $4n+1 < (\sqrt{n} + \sqrt{n+1})^2 < 4n+3$.