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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.

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  • $\begingroup$ We need that $p\ge 1$, since otherwise the square root is not defined. That is also a sufficient condition. $\endgroup$ – Klaas van Aarsen Dec 12 '18 at 17:48
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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,

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

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

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