Find the invertible solutions of $x^{-1}+x=0$ over a ring.

Question

Let $(R, +, \times)$ be a finite ring with identity $1$. There are two notations:The set of all invertible elements in $(R, \times)$ and the set of all nonzero elements in $R$, are denoted by $R^\times$ and $R^*$, respectively.

I am considering the following problem.

Solve the equation over $R^\times$: $x^{-1}+x=0$.

I have tried the following:

Suppose the multiplicative order of $e$ is $r$, denoted by $r=\text{ord}(e)$. It means that $x^r=1$. Then $x^{-1}=x^{r-1}$. The equation now is $x^{r-1}+x=0$.

Since $x$ is invertible, the equation now is $x^{r-2}+1=0$.

Then it has $x^{r-2}=-1$. Squaring both sides, we get $x^{2(r-2)}=1$ which implies that $r\mid 2(r-2)$. Equivalently, $4\mid r$.

Case 1: $R$ is a finite field $\mathbb{F}_q$. Then $-1=g^{\frac{q-1}{2}}$ and $x=g^i$ for some $1\le i\le q-1$, where $g$ is a generator of $\mathbb{F}_q^\times$. The equation now is $g^{i(r-2)}=g^{\frac{q-1}{2}}$. That is $i(r-2)\equiv \frac{q-1}{2}\pmod{q-1}$. Then it has solutions for $i$, if and only if, $\frac{q-1}{2}\mid\gcd(r-2, q-1)$. And then $$\gcd(r-2, q-1)=\begin{cases}q-1 \\ \frac{q-1}{2} \end{cases}$$ If $\gcd(r-2,q-1)=q-1$, then $(q-1)\mid (r-2)$. it is a contradiction since $r \mid (q-1)$. If $\gcd(r-2,q-1)=\frac{q-1}{2}$, then we have $\frac{q-1}{2}\mid (r-2)$. Let $k_1\ge0$ and $k_2\ge 0$ be two integers such that $$r-2=k_1\frac{q-1}{2}, q-1=k_2r.$$ We have $(2-k_1k_2)r=4=1\times 4=2\times 2 =4\times 1$. To solve the equation of two variables $k_1,k_2$ and $r$, we get two solutions:(If $r=1$, then $k_1k_2=-2$ which is a contradiction.)

1. $r=4, k_1=k_2=1$
2. $r=2,k_1=0$ and $2k_2+1$ is a prime power.

To sum up， we have $$\{x\in \mathbb{F}_q^\times \mid x^{-1}+x=0\}=\begin{cases} \{2,3\}, & q=5\\ \{1,-1\}, &q=2k+1, \end{cases}$$

Case 2: $R$ is a residual class ring $\mathbb{Z}_n$. No ideas....

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PS: the trying over finite field $\mathbb{F}_q$ is failed after I run the following Magma code：

for q in [2..1000] do
    if not IsPrimePower(q) then
        continue;
    end if;
    fq:=GF(q);
    solutions:={x: x in (Set(fq) diff {0}) | ((x)^(-1)+(x)) eq 0};
    if #solutions gt 0 then
        print q, #solutions, solutions;
    end if;
end for;

The code gives many counterexample such as, for $q=13$, the solutions should be $5$ and $8$. And the numeric experiment evidences that

1. When $2\mid q$, the only one solution would be $x=1$;
2. Otherwise, it has exactly two solutions $x_1$ and $x_2$ such that $x_1+x_2=0$.

Now I have no ideas at this moment. Maybe it is hard to answer this problem. It would be easier if the ring is residual class ring $\mathbb{Z}_n$ or finite field $\mathbb{F}_q$.

Or, there is a posibility that the problem was well-studied and the result is unknown for me. Then please give me some keywords or conferences.

Thanks for any replies.