Prove that $\lfloor \sqrt{(p-1)p} \rfloor = p - 1$ and likewise $\lceil \sqrt{(p-1)p} \rceil = p$.

Question

Here $p$ is prime but is not necessary for the problem just that $p \ge 0$. I suspect that a statement like $p-1 \le \sqrt{(p-1)p} \le p$ would be the case but I am not certain how to establish this condition.

Answer 1

For $p>1$,$$(p-1)^2=p^2-2p+1<p^2-p<p^2$$ and

$$p-1<\sqrt{(p-1)p}<p.$$

As the extreme members are integer,

$$p-1\le\left\lfloor\sqrt{(p-1)p}\right\rfloor\le\left\lceil\sqrt{(p-1)p}\right\rceil\le p.$$

Answer 2

Hint: $(p-1)^2 \le (p-1)p \le p^2$ for $p \ge 0$