Fixed: Is this set empty? $ S = \{ x \in \mathbb{Z} \mid \sqrt{x} \in \mathbb{Q}, \sqrt{x} \notin \mathbb{Z}, x \notin \mathbb{P}$ } This question has been "fixed" to reflect the question that I intended to ask
Is this set empty? $ S = \{ x \in \mathbb{Z} \mid \sqrt{x} \in \mathbb{Q}, \sqrt{x} \notin \mathbb{Z}, x \notin \mathbb{P}$ }
Is there a integer, $x$, that is not prime and has a root that is not irrational?

To answer according to the last edit: Yes.
Let $a\in\mathbb{Z}$ and consider the polynomial $x^{2}-a$. 
By the rational root theorm if there is a rational root $\frac{r}{s}$
then $s|1$ hence $s=\pm1$ and the root is an integer. So $\sqrt{a}\in\mathbb{Q}\iff\sqrt{a}\in\mathbb{Z}$ .
Since you assumed that the root is not an integer but is a rational number this can not be.
