2018/005) Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+1}}\right\rfloor$

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