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.
Best Answer
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): $$ \begin{array}{|c|c|} \hline n & \alpha^n\\ \hline 0 & 1\\ 1 & \alpha\\ 2 & 3\alpha + 2\\ 3 & \alpha + 1\\ 4 & 4\alpha + 2\\ 5 & 4\alpha + 3\\ 6 & 3\\ 7 & 3\alpha\\ 8 & 4\alpha + 1\\ 9 & 3\alpha + 3\\ 10 & 2\alpha + 1\\ 11 & 2\alpha + 4\\ 12 & 4\\ 13 & 4\alpha\\ 14 & 2\alpha + 3\\ 15 & 4\alpha + 4\\ 16 & \alpha + 3\\ 17 & \alpha + 2\\ 18 & 2\\ 19 & 2\alpha\\ 20 & \alpha + 4\\ 21 & 2\alpha + 2\\ 22 & 3\alpha + 4\\ 23 & 3\alpha + 1\\ \hline \end{array} $$