Suppose $g$ is a diagonalizable element of $SL_2(\mathbb{F}_p)$ (we call an element of $SL_2(\mathbb{F}_p)$ diagonalizable, iff it has two linearly independent eigenvectors). Then the order of $g$ is the Least Common Multiple of orders of its eigenvalues $e_1$ and $e_2$ (they lie in the algebraic closure of $\mathbb{F}_p$, "order of a field element $x$" stands for the minimal natural $n$, such that $x^n = 1$). However, as $det(g) = 1$, $e_1e_2 = 1$. That means, that orders of the eigenvalues of $g$ are equal (and thus equal to the order of $g$). Also, it is easy to notice, that $x$ is an eigenvalue of $g$ iff it satisfies the equality $x^2 - tr(g)x + 1 = 0$. So one can conclude, that the order of any diagonalizable element of $SL_2(\mathbb{F}_p)$ depends solely on its trace. So there exists a function $f$ from $\mathbb{F}_p$ to $\mathbb{N}$, such that $f(x)$ is order of any diagonalizable element of $SL_2(\mathbb{F}_p)$ with trace $x$. The question is: how to write the function $f$ in a closed form?
Any help will be appreciated.