2014/102) show that $\lfloor(\sqrt{n} +\sqrt{n+1}\rfloor= \lfloor(\sqrt{4n+2}\rfloor$

we realise that $n(n+1) = (n+\dfrac{1}{2})^2 - \dfrac{1}{4}$
so $\sqrt{n(n+1)}\lt(n+\dfrac{1}{2})$
clearly $n\lt\sqrt{n(n+1)}$

so

we have
$(\sqrt{n} + \sqrt{n+1})^2 = n + n+ 1 + 2 \sqrt{n(n+1)}$
= $2n +1 + 2 \sqrt{n(n+1)}$
>$2n+ 1 + 2 n$ or > $4n+ 1$

and = $2n +1 + 2 \sqrt{n(n+1)}$ <  $2n +1 + 2 (n + \dfrac{1}{2})$
or <  (4n + 2)
so $(4n+1)\lt(\sqrt{n} + \sqrt{n+1})^2\lt(4n+2)$

because  4n+2 is not a perfect square

we have $\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor$
and as $(\sqrt{n} + \sqrt{n+1})^2$ is between 4n + 1 and 4n +2 we have

$\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor= \lfloor(\sqrt{n} +\sqrt{n+1}\rfloor$