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According to an answer in this question the set of rational non-squares is diophantine over the rationals: there is polynomial $P(a,x_1,\dots,x_n)$ which for $a\in\mathbb{Q}$ has a rational solution iff $a$ is a rational non-square. For the variety take the hypersurface with equation $P^2+(x_{n+1}^2-a)^2=0$ in $n+2$ coordinates $(a,x_1,\dots,x_{n+1}$).