# Understanding this proof regarding quadratic residues

Let $$p$$ be an odd prime and let $$Q_p$$ denote the quadratic residues modulo $$p$$, $$N_p$$ the non-residues modulo $$p$$.

Let $$X$$ be some subset of $$p$$. Then, $$q X \equiv X mod (p) \hspace{2mm}\forall q \in Q_p\iff X \textit{is the union of {\{0}\}, Q_p, or N_p}$$

Now, I'm struggling to see why this statement is true.

For $$q \in Q_p$$ and $$r \in N_p$$, I know that $$Q_{p} q = Q_p, Q_p r = N_p, N_p q = N_p, N_p r = Q_p$$, this seems to be useful information regarding the proof of the statement. This information clearly proves the left direction.

How to prove the ($$\implies$$) direction?

Hint: Observe that, supposed $$q$$ is nonzero quadratic residue, $$qx$$ will be a qu. residue if and only if $$x$$ is.
Thus, for a nonzero $$x\in X$$, the set $$\{qx\mid q\in Q_p\}$$ has $$|Q_p|$$ elements modulo $$p$$.