Can you prove this floor function challenge involving square roots?

[anemone]

Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor$ for any positive integer $n$.

 

[kaliprasad]

Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor$ for any positive integer $n$.

Spoiler

LHS = $\lfloor{\sqrt{n}+\sqrt{n+1}}\rfloor $
= $\lfloor\sqrt{(\sqrt{n}+\sqrt{n+1})^2}\rfloor $
= $\lfloor\sqrt{n+n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{(n+\dfrac{1}{2})^2 + \dfrac{3}{4}}}\rfloor $
= $\lfloor\sqrt{2n+1+\sqrt{(2n+1)^2 + 3}}\rfloor $

now realising that integral part of square root of x and square root of x + t where t is less than 1 are sameso we need to find the integral part of $\sqrt{(2n+1)^2 + 3}$

$(2n+1)^2 + 3\gt(2n+1)^2$
and $(2n+1)^2 + 3\lt(2n+2)^2$ as $(2n+2)^2-(2n+1)^2 = 4n + 3$so integral part of $\sqrt{(2n+1)^2 + 3}$ = (2n + 1)

so LHS = $\lfloor\sqrt{2n+1+2n+1}\rfloor $
= $\lfloor\sqrt{4n+2}\rfloor $
= RHS

 

[anemone]

Spoiler

LHS = $\lfloor{\sqrt{n}+\sqrt{n+1}}\rfloor $
= $\lfloor\sqrt{(\sqrt{n}+\sqrt{n+1})^2}\rfloor $
= $\lfloor\sqrt{n+n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{n(n+1)}}\rfloor $
= $\lfloor\sqrt{2n+1+2\sqrt{(n+\dfrac{1}{2})^2 + \dfrac{3}{4}}}\rfloor $
= $\lfloor\sqrt{2n+1+\sqrt{(2n+1)^2 + 3}}\rfloor $

now realising that integral part of square root of x and square root of x + t where t is less than 1 are sameso we need to find the integral part of $\sqrt{(2n+1)^2 + 3}$

$(2n+1)^2 + 3\gt(2n+1)^2$
and $(2n+1)^2 + 3\lt(2n+2)^2$ as $(2n+2)^2-(2n+1)^2 = 4n + 3$so integral part of $\sqrt{(2n+1)^2 + 3}$ = (2n + 1)

so LHS = $\lfloor\sqrt{2n+1+2n+1}\rfloor $
= $\lfloor\sqrt{4n+2}\rfloor $
= RHS

Well done and thanks for participating, kaliprasad!(Yes)

Here's another solution that I want to share with MHB:

Spoiler

It's easy to verify that

$\sqrt{4n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4n+3}$ where $n\in N$

Neither $4n+2$ nor $4n+3$ are squares, so $\left\lfloor{\sqrt{4n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+2}}\right\rfloor=\left\lfloor{\sqrt{4n+3}}\right\rfloor$ and the result follows.