Existence of a quadratic residue $r$ mod $p$ such that $-r-1$ is a quadratic nonresidue.

Question

I would be able to simplify a proof about sums of squares significantly if I were able to prove that the involution $$\Bbb{F}_p\ \longrightarrow\ \Bbb{F}_p:\ x\ \longmapsto\ -x-1,$$ does not preserve quadratic residues. I've checked a few small examples and it seems to hold, and it seems less likely to fail as $p$ grows, but I'm unable to prove it. Is this true, and if yes, what is a nice way to see this?

Answer 1

Irreducible conics over $\mathbb{F}_q$ have $q+1$ points in $\mathbb{P}^{2}(\mathbb{F}_q)$. The line $z=0$ intersects $x^2+y^2+z^2=0$ at the two points $(i:1:0)$, where $i^2=-1$, and these points are defined over $\mathbb{F}_q$ iff $q\equiv 1 \mod 4$. That is, the affine curve $$x^2+y^2+1=0$$ has either $q-1$ points or $q+1$ points in $\mathbb{A}^{2}(\mathbb{F}_q)$. For simplicity, we'll say that $x^2+y^2+1=0$ has about $q$ solutions.

Now note that if $(a,b)$ is a solution, then so is $(- a, b)$. Thus, for $S=\{i^2: i \in \mathbb{F}_q \}$, the equation $$y^2=-x-1,$$ has about $q/2$ solutions $(x,y) \in S\times \mathbb{F}_q $. Since such solutions come in pairs $(x_0,\pm y_0)$, and $S$ has about $q/2$ elements, the equation $$y^2=-x-1,$$ will be solvable for about half of the elements $x\in S$.

Answer 2

Let $p$ be an odd prime.

The goal is to show that the map

$$\Bbb{F}_p\ \longrightarrow\ \Bbb{F}_p:\ x\ \longmapsto\ -x-1$$

does not preserve quadratic residues mod $p$.

Suppose $p = 1 \pmod 4$. Then $-1$ is not a quadratic residue mod $p$. Since you stated that you are regarding $0$ as a quadratic residue mod $p$, the proof for this case is immediate since $0$ maps to $-1$.

Even if we remove $0$ and $-1$ from consideration, then for p > 3, as shown below, it will still be the case that the map in question does not preserve quadratic residues mod $p$.

Proposition:

If $p$ is prime, $p > 3$, there exists $x \in Z_p$, $0 < x < p-1$ such that $x$ is a quadratic residue but $-x - 1$ is not a quadratic residue.

Proof:

Assume $p$ is prime, $p > 3$, and let $f\,\colon Z_p \rightarrow Z_p$ be defined by $f(x)=-x - 1$.

Let $A=\{0,p-1\}$, $B=\{1,...,p-2\}$.

Note that $f(0)=-1$ and $f(-1) = 0$, hence $f(A) = A$.

Clearly, $f$ is a bijection from $Z_p$ to $Z_p$.

Since $f(A)=A$, it follows that $f(B)=B$, so $f$ restricted to $B$ is a bijection from $B$ to $B$.

To prove the proposition, it suffices show that $f$ restricted to $B$ does not preserve quadratic residues mod $p$.

Let $S=\{x \in B \mid x \text{ is a quadratic residue mod } p\}$, and let $T=\{f(x) \mid x \in S\}$. Claim $S \ne T$.

The nonzero quadratic residues mod $p$, are the roots in $Z_p$ of $x^{(p-1)/2}-1$. Since $p > 3$, Viete's formula implies that the sum in $Z_p$ of the nonzero quadratic residues is $0$.

Consider two cases, according as $p \equiv -1 \pmod 4$ or $p \equiv 1 \pmod 4$.

Case $(1)\,$: $\;\;p \equiv -1 \pmod 4$.

Then $-1$ is not a quadratic residue mod $p$, hence $S$ is the set of all nonzero quadratic residues mod $p$.

It follows that, reduced mod $p$, $\displaystyle{\sum_{x \in S} x = 0}$.

However, reduced mod $p$,

\begin{align*} \sum_{y \in T} y &= \sum_{x \in S} f(x)\\[6pt] &= \sum_{x \in S} (-x-1)\\[6pt] &= -\bigg(\sum_{x \in S} x\bigg) - \frac{p-1}{2}\\[6pt] &= -\frac{p-1}{2}\\[6pt] &\ne 0\\ \end{align*}

hence, for case $(1)$, $S \ne T$.

Case $(2)\,$: $\;\;p \equiv 1 \pmod 4$.

Then $-1$ is a quadratic residue mod $p$, hence $S$ is the set of all nonzero quadratic residues mod $p$, other than $-1$.

It follows that, reduced mod $p$, $\displaystyle{\sum_{x \in S} x = 1}$.

However, reduced mod $p$,

\begin{align*} \sum_{y \in T} y &= \sum_{x \in S} f(x)\\[6pt] &= \sum_{x \in S} (-x-1)\\[6pt] &= -\bigg(\sum_{x \in S} x\bigg) - \frac{p-1}{2}\\[6pt] &= -1- \frac{p-1}{2}\\[6pt] &= -\frac{p+1}{2}\\[6pt] &\ne 1\;\;\;\text{[since $p \ne 3$]}\\ \end{align*} Hence, for case $(2)$, $S \ne T$, and so in both cases, $S \ne T$.

It follows, as claimed, that $f$ restricted to $B$ does not preserve quadratic residues mod $p$.

This completes the proof of the proposition.