Quadratic residues - invariant under translation Let $p$ be an odd prime number. 
Can the set of squares modulo $p$ be invariant under translation? 
I.e. given $p$, let $S = (\mathbb{F}_p^\times)^2 \cup \{0\} \subseteq \mathbb{F}_p$. Can there exist $\delta \in \mathbb{F}_p^\times$ such that $$S + \delta := \{x + \delta \mid x \in S\}$$ is again equal to $S$?
I suspect that the answer is no...
 A: Assume by contradiction that $S=S+\delta$ for some $\delta \neq 0$.
Then, as $0 \in S$ we get $\delta \in S$ and then by induction that $n\delta \in S$.
But the additive group $(\mathbb F_p,+)$ is cyclic and generated by any non-zero element. Thus 
$$F_p=\{ n \delta | z \in \mathbb N \} \subset S \,.$$
Contradiction.
A: No: The squares of $\mathbf F_p$ are the roots of
$$f(X) = X^{\frac{p+1}{2}} - X.$$
If $\delta$ is as you describe, then
$$f(X) = f(X+\delta).$$
Remark that $$f'(X) = \frac{p+1}{2}X^{\frac{p-1}{2}} - 1.$$
Differentiating the equation $f(X) = f(X+\delta)$ we find
$$\frac{p+1}{2}X^{\frac{p-1}{2}} - 1 = \frac{p+1}{2}(X+\delta)^{\frac{p-1}{2}} - 1.$$
Putting $X=0$ we see that $\delta^{\frac{p-1}{2}} = 0$, so $\delta =0$.
A: You are right, the answer is no, but it has little to do with quadratic residues.
First note that if a set $S$ is invariant under translation in $\mathbb{F}_p$ it means that either $S$ is empty or $S$ is the whole $\mathbb{F}_p$. In our case $S$ is clearly not empty as $0$ and $1$ are always quadratic residues.
If $a \in S$, then clearly $\{a + \delta,\ a+2\delta,\ a+3\delta,\ \dots\} \subseteq S$, since $S$ is invariant under translation by $\delta$, but note that for any $\delta \in \mathbb{F}_p \setminus \{0\}$ the subgroup generated by $\delta$ must be the entire $\mathbb{F}_p$, since the order of a subgroup divides the order of $\mathbb{F}_p$ which is $p$.
So $\{\delta,\ 2\delta,\ 3\delta,\ \dots\} = \mathbb{F}_p \Longrightarrow \{a+\delta,\ a+2\delta,\ a+3\delta,\ \dots\} = \mathbb{F}_p \Longrightarrow S \supseteq \mathbb{F}_p \Longrightarrow  S = \mathbb{F}_p$
But quadratic residues cannot be the whole $\mathbb{F}_p$ since $a^2 \equiv (p-a)^2$ limits their number enough for $p \geq 3$ ($-1 \neq\ 1$, but $(-1)^2 \equiv 1^2$). So there's no prime and no shift which can produce the same set of quadratic residues.
