Let $n\in \mathbb N,$ show that $\lfloor \sqrt{4n+2}\rfloor=\lfloor \sqrt{4n+1}\rfloor$ and $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor \sqrt{4n+1}\rfloor$.
For $n\in \mathbb N$, let $p=\lfloor \sqrt n \rfloor $, then we have : $$p \leqslant \sqrt n <p+1,$$ so $p^2 \leqslant n < (p+1)^2$ and $p^2+1 \leqslant n+1 < (p+1)^2+1$ Then $$p \sqrt{1+1/p^2} \leqslant \sqrt{n+1} \leqslant (p+1)\sqrt{1+1/(p+1)^2} $$ The prolem is that I can't deduce that $$p \leqslant \sqrt {n+1} \leqslant p+1 $$ Any help is really appreciated !
