JonnyV wrote: Prove that $\lfloor{\sqrt{9n+7}}\rfloor=\lfloor{\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}}\rfloor$ for all postive integer $n$. $\sqrt n+\sqrt{n+2}<2\sqrt{n+1}$ $\implies$ $\sqrt n+\sqrt{n+1}+\sqrt{n+2}<\sqrt{9n+9}$ $\sqrt n+\sqrt{n+2}>\sqrt{4n+3}$ $\sqrt{4n+3}+\sqrt{n+1}>\sqrt{9n+7}$ And so $\sqrt n+\sqrt{n+1}+\sqrt{n+2}>\sqrt{9n+7}$ So $\sqrt{9n+7}<\sqrt n+\sqrt{n+1}+\sqrt{n+2}<\sqrt{9n+9}$ And since $9n+8$ is never a perfect square, we got the required result.