primitive element $\alpha$ in $\mathbb{F}_{25}$ Find a primitive element $\alpha$ in $\mathbb{F}_{25}$ and for every $\beta \in \mathbb{F}_{25}^*$ find the least $n\in \mathbb{Z}^+$ such that $\alpha^n=\beta$. 
I constructed $\mathbb{F}_{25}$ by $\mathbb{F}_{5} / (x^2+2x+3)$ but I am not sure how to find a primitive element as there are $25$ orders to compute. I tried $\alpha$ as a root of the polynomial I used, and I got $\alpha^3=1$ so obviously that is not a generator $\mathbb{F}_{25}^*$.
 I found another construction by $\mathbb{F}_{5} / (x^2+4x+2)$  and then the root of this polynomial, say $\alpha$, in $\mathbb{F}_{5} / (x^2+2x+3)$ is primitive but I would like to know how to do this for my construction.

If $\alpha \in \mathbb F_{25}$ has minimal polynomial $P$ over $\mathbb F_5$, then $\alpha$ is a primitive element iff $P$ divides the $24$th cyclotomic polynomial over $\mathbb F_5$.
So to prove that $\alpha$ is a primitive element, it suffices to check that $x^2+2x+3$ does not divide $\Phi_1 = x-1$, $\Phi_2=x+1$, $\Phi_3 = x^2+x+1$, $\Phi_4 = x^2+1$, $\Phi_6 = x^2+4x+1$, $\Phi_8 = x^4+1 = (x^2+2)(x^2+3)$, $\Phi_{12} = x^8-x^4+1 = (x^2+2x+4)(x^2+3x+4)$.
