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.
 A: 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 $\Phi_1 = x-1$, $\Phi_2=x+1$, $\Phi_3 = x^2+x+1$, $\Phi_4 = x^2+1$, $\Phi_6 = x^2+4x+1$, $\Phi_8 = x^4+1 = (x^2+2)(x^2+3)$, $\Phi_{12} = x^8-x^4+1 = (x^2+2x+4)(x^2+3x+4)$.
A: When you say $\Bbb F_5/(x^2+2x+3)$, I think you ought to write $\Bbb F_5[x]/(x^2+2x+3)$. Because that's what I think you mean.
If $\alpha$ is a root of your polynomial, then we have $\alpha^2 = 3\alpha + 2$, and $$\alpha^3 = \alpha^2\cdot \alpha = (3\alpha +2)\alpha\\
= 3\alpha^2 +2\alpha = 3(3\alpha + 2) + 2\alpha\\
= \alpha + 1\neq 1$$
Here is a full table of the powers of $\alpha$, generated by a rather simple Python script (the print format may look  strange, but it's specifically constructed to make it copy-pastable into the below table, with minimal need for tidying up):
$$
\begin{array}{|c|c|}
\hline n & \alpha^n\\
\hline 0 & 1\\
1 & \alpha\\
2 & 3\alpha + 2\\
3 & \alpha + 1\\
4 & 4\alpha + 2\\
5 & 4\alpha + 3\\
6 & 3\\
7 & 3\alpha\\
8 & 4\alpha + 1\\
9 & 3\alpha + 3\\
10 & 2\alpha + 1\\
11 & 2\alpha + 4\\
12 & 4\\
13 & 4\alpha\\
14 & 2\alpha + 3\\
15 & 4\alpha + 4\\
16 & \alpha + 3\\
17 & \alpha + 2\\
18 & 2\\
19 & 2\alpha\\
20 & \alpha + 4\\
21 & 2\alpha + 2\\
22 & 3\alpha + 4\\
23 & 3\alpha + 1\\
\hline
\end{array}
$$
