Gazeta Matematica 6-7-8/2016, Problem 27246

Problem:
Do there distinct prime numbers $p,q,r$ such that $\sqrt{p},\sqrt{q},\sqrt{r}$ are terms of an arithmetic progression?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
The answer is no. Assume by contradiction that $\sqrt{p},\sqrt{q},\sqrt{r}$ are in the same arithmetic progression. If $d \neq 0$ is the common difference, then there exist integers $m,n \neq 0$ such that $$\begin{array}{rcl}\sqrt{q}-\sqrt{p}&=&md, \\ \sqrt{r}-\sqrt{p}&=&nd. \end{array}$$ Dividing the first and the second equation by $m$ and $n$ respectively and substituting, we obtain $$m(\sqrt{r}-\sqrt{p})=n(\sqrt{q}-\sqrt{p}),$$ whence $$m\sqrt{r}-n\sqrt{q}=(m-n)\sqrt{p}.$$ Squaring both sides, we obtain $$rm^2+qn^2-2mn\sqrt{rq}=p(m-n)^2,$$ which gives $$\sqrt{rq}=\dfrac{rm^2+qn^2-p(m-n)^2}{2mn}.$$ But this equality is false, since the left-hand side is irrational and the right-hand side is rational.