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 $p$.


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