Let $QR_p \subseteq \mathbb{F}_p^*$ be the set of quadratic residues modulo $p$.
Is there a prime $p$ and an integer $x \not = 0$ mod $p$ such that $QR_p$ is invariant under addition with $x$?
i.e. $QR_p = QR_p + x := \{(q+x)~mod~p~|~q \in QR_p\}$
I think the answer should be no. If the answer was yes, there would be an arithmetic progression with common difference $x$, but I don't know how to prove/disprove the statement.
If $0$ is considered a quadratic residue mod $p$: No such $x$ exists. Note that $0\in QR_p$, so $x+0\equiv q^2\pmod{p}$ implies that $x$ is a quadratic residue. In order for $QR_p+x$ to be equal to $QR_p$, we would need $nx$ to be a quadratic residue mod $p$ for every $n\in\mathbb{Z}$; clearly this is not true, since if take $n$ to be any number which is not a quadratic residue mod $p$, then $nx\notin QR_p$.
If $0$ is not considered a quadratic residue mod $p$: No such $x$ exists. Note that for fixed $m\in QR_p$, we would need $nx-m\equiv 0\pmod{p}$ for every $n\in\mathbb{Z}$. In particular, taking $n=m$ yields: $m(x-1)\equiv0\pmod{p}$, so $p|m$ (not true) or $p|(x-1)$, which would mean that $x$ has to be $1$. Now, let $n$ be a not quadratic residue mod $p$. Then $(n-1)mx+m=nm$ is not a quadratic residue mod $p$.