What would be the conditions (if any) on the trace of an element in $SL(2,p)$ in order for it to have order 5 ? (assuming $p= \pm1 mod 10$)
For example, any traceless element in $SL(2,p)$ has order 4 (straightforward proof).
Any suggestion or comment is tremendously valuable.
Let's look at an example. Let $p=11$. The primitive 5th roots (modulo 11) are $3,4,5,9$, where $3$ and $4$ are inverse, and $5$ and $9$ are inverse. So the trace has to be $3+4=7$ or $5+9=3$.
These necessary conditions are also sufficient. If the determinant is $1$ and the trace is $7$, then the eigenvalues must be $3$ and $4$, so the order must be 5; similarly if the trace is 3.
Maybe this will give some idea how to proceed in the general case.
Consider $G=\text{SL}(2,F)$ for an arbitrary field $F$. For an element to have order $m$ (coprime to the characteristic of $k$ if this is prime) its eigenvalues are $\zeta$ and $\zeta^{-1}$ where $\zeta$ is a primitive $m$-th root of unity, that is $\zeta$ is a zero of the $m$-th cyclotomic polynomial $\Phi_m$. As long as $m\ge3$, the trace $t=\zeta+\zeta^{-1}$ must therefore be a root of the "real cyclotomic" polynomial $\Psi_m$. This isn't an established name or notation, but is the minimal polynomial for $2\cos(2\pi/m)$ and satisfies $$\Psi_m(X+X^{-1})=X^{-\phi(m)/2}\Phi_m(X)$$ where $\phi$ is Euler's totient. For example $$\Psi_5(X+X^{-1})=X^2+X+1+X^{-1}+X^{-2}$$ and so $$\Psi_5(Y)=Y^2+Y-1.$$
Suppose $A$ is an element of order $5$ in $\DeclareMathOperator{\SL}{SL} \SL_2(\mathbb{F}_p)$ and let $m(x)$ be its minimal polynomial. If $m$ has degree $1$, then $A$ is a scalar matrix hence must be of the form $$ \begin{pmatrix} c & 0\\ 0 & c \end{pmatrix} $$ for some $c$. But then $c^2 = \det(A) = 1$, so $A$ has order at most $2$, contradiction. Thus $m$ must have degree $2$, so we can write $m(x) = x^2 + ax + 1$ for some $a$. Since $A$ satisfies $x^5 - 1$, then we must have that $x^2 + ax + 1$ divides $x^5 - 1$.
First, assume that $x-1 \nmid x^2 + ax + 1$. Then $x^2 + ax + 1$ divides $x^4 + x^3 + x^2 + x + 1$. Using division with remainder, we find that \begin{align*} x^4 + x^3 + x^2 + x + 1 &= (x^2 + (1-a)x + a(a-1))m(x) + (-a^3 + a^2 + a)x + (-a^2 + a + 1) \end{align*} Thus we must have that $a$ satisfies $a^2 - a - 1 = 0$. The quadratic equation $t^2 - t - 1$ has discriminant $5$, hence has a solution iff $5$ is a square mod $p$. Since $(5|p) = (-1)^{p-1} (p|5) = (p|5)$ by quadratic reciprocity and the only squares mod $5$ are $0,1,4$, then we can find such an $a$ iff $p=5$ or $p \equiv 1, 4 \pmod{5}$. (Here $(\cdot | \cdot)$ denotes the Legendre symbol.)
In this case, then $a = \frac{1 \pm \sqrt{5}}{2}$ so $$ \begin{pmatrix} 0 & -1\\ 1 & -\frac{1 \pm \sqrt{5}}{2} \end{pmatrix} $$ is an element of order $5$. (This is the companion matrix for $x^2 + ax + 1$ for the two possible values of $a$. In fact, by uniqueness of rational canonical form, every such matrix must be similar to one of these matrices.)
If $x-1 \mid x^2 + ax + 1$, then $x^2 + ax + 1 = x^2 - 2x + 1$. Again using polynomial division, we find $$ x^5 - 1 = (x^3 + 2x^2 + 3x + 4)(x^2-2x+1) + 5x-5 $$ so we must have $5 = 0$ in $\mathbb{F}_p$, i.e., $p=5$. (Thus we discover no new primes in this latter case.) In this case, $a = -2 = 3$ is also equal to $\frac{1 \pm \sqrt{5}}{2} = \frac{1}{2}$.
Thus in all cases, we find that $\operatorname{Tr}(A) = -a = -\frac{1 \pm \sqrt{5}}{2}$.
For instance, for $p=11$ we have $4^2 = 16 = 5$ and $$ \frac{1 \pm \sqrt{5}}{2} = \frac{1 \pm 4}{2} = \frac{5}{2}, \, \frac{8}{2} = 8, 4 $$ are the roots of $t^2 - t - 1$. Then $$ \begin{pmatrix} 0 & -1\\ 1 & -4 \end{pmatrix} \qquad \begin{pmatrix} 0 & -1\\ 1 & -8 \end{pmatrix} $$ have order $5$.
