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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...

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For, an immediate necessary condition is that both $\delta$ and $-\delta$ be squares, i.e. $p\equiv1\bmod4$. – Andrea Mori Feb 27 '14 at 13:06
up vote 4 down vote accepted

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.

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Hmmm... how did I not see this! This is much better. – Bruno Joyal Feb 27 '14 at 13:25

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 $\delta =0$.

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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.

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