Cfrgtkky

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 $rmid 2(r-2)$. Equivalently, $4mid r$.

Case 1: $R$ is a finite field $mathbb{F}_q$. Then $-1=g^{frac{q-1}{2}}$ and $x=g^i$ for some $1le ile q-1$, 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, $frac{q-1}{2}midgcd(r-2, q-1)$. 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 $k_1ge0$ and $k_2ge 0$ be two integers such that
$$r-2=k_1frac{q-1}{2}, q-1=k_2r.$$
We have $(2-k_1k_2)r=4=1times 4=2times 2 =4times 1$. 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
$${xin mathbb{F}_q^times mid x^{-1}+x=0}=begin{cases}
{2,3}, & q=5\
{1,-1}, &q=2k+1,
end{cases}$$

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

=====================================================

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 $2mid q$, 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.