An inefficient way to solve these equations is to note that you are working over . Therefore, there are finitely many cases to check. You can start with and go to (for those are elements in the finite field) and see if you get any solutions. For example, say . The first equation forces which in particular means are multiplicative inverses. Now look at second equation, since by assumption it means , but is invertible and so . Therefore, the third equation gives nothing and and the fourth equation also does not tells us anything new. Thus, all we have and . Therefore, if then the matrices that have this property have and the entries on such that . These include, . This contributes elements. Now move over to the case and so on. Those cases get more involved but the same idea applies.

As for the hint the theorem goes as follows: let be a finite group so that is divisible by then the number of solutions to is a multiple of . However, there is one solution which is trivial, that is . That is the only solution that has order . While all other solutions have order . Thus, if is the number of elements of order then it follows that .