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Actually, it's better to link directly to the specific answer, in case more will come in the future. Also, give credit where credit is due.
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Emil Jeřábek
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According to an answer in this questionthis answer by Laurent Moret-Bailly, the set of rational non-squares is diophantine over the rationals (a result of Bjorn Poonen): there is a 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 $(a,x_1,\dots,x_{n+1}$)$(a,x_1,\dots,x_{n+1})$.

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 $(a,x_1,\dots,x_{n+1}$).

According to this answer by Laurent Moret-Bailly, the set of rational non-squares is diophantine over the rationals (a result of Bjorn Poonen): there is a 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 $(a,x_1,\dots,x_{n+1})$.

fix one more integer, and fix link so that it does not go to an irrelevant comment
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Emil Jeřábek
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According to an answer in this questionthis question the set of integerrational non-squares is diophantine over the rationals: there is polynomial $P(a,x_1,...,x_n)$$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 $(a,x_1,\dots,x_{n+1}$).

According to an answer in this question the set of integer non-squares is diophantine over the rationals: there is polynomial $P(a,x_1,...,x_n)$ 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 $(a,x_1,\dots,x_{n+1}$).

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 $(a,x_1,\dots,x_{n+1}$).

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

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

According to an answer in this question the set of integer non-squares is diophantine over the rationals: there is polynomial $P(a,x_1,...,x_n)$ 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 $(a,x_1,\dots,x_{n+1}$).

added equality
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joro
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added 19 characters in body
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joro
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joro
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