Inspired by the Eliptic Curve Primality Test, and classical primality tests, I wanted to know if any particular equation (using multivariate polynomials) over finite fields.
The group $(\mathbb Z/n\mathbb Z)^{*}$ is the multiplicative group of integers $\pmod n$. It is known that $|(\mathbb Z/n\mathbb Z)^{*}| = n-1$. The classical primality test is as follows:
If there exists an integer $a$ such that $a^{n-1}=1\pmod n$ and $a^{(n-1)/q}≠1\pmod n$ for each prime $q | n-1$, then $n$ is prime. This follows from the fact that $a$ is a generator of $(\mathbb Z/n\mathbb Z)^{*}$, and the group is cyclic. Example: $n=17$:
$a=3$, $3^{16}=1\pmod {17}$, $3^8=16≠1\pmod {17}$. Since $2$ is the only prime dividing $n-1=16$, the order of $3\pmod {17}$ must be 16, and $17$ is indeed prime.
What if instead of using $(\mathbb Z/n\mathbb Z)^{*}$, we used curves or equations over finite fields, C$ \mathbb F(n)$, and thus formed a group C over the multiplicative group $(\mathbb Z/n\mathbb Z)^{*}$?
Suppose $C$ is the equation formed by $y^2 = x^4 - 2x^3 + x - 1 \pmod n$ (this is very similar to elliptic curves, I wanted to use quartics instead to be creative). Using $n=17$, we first find all solutions to $y^2 = x^4 - 2x^3 + x - 1 \pmod {17}$. There are 20 of them:
$(0, 4), (0, 13), (1, 4), (1, 13), (2, 1), (2, 16), (6, 6), (6, 11), (7, 2), (7, 15), (8, 6), (8, 11), (10, 6), (10, 11), (11, 2), (11, 15), (12, 6), (12, 11), (16, 1), (16, 16)$
Do the same principles for finding the cardinality of such groups apply here? There are $20$ points listed before the point of infinity, so the cardinality of this group is |C$ \mathbb F(17)$| = $21$? What is the cardinality of the multiplicative group, $|C (\mathbb Z/17\mathbb Z)^{*}|$?
Is there a known algorithm to find a generator (one of the $20$ elements) of $C (\mathbb Z/17\mathbb Z)^{*}$ or C$ \mathbb F(17)$ similar to how $3$ is a generator of the multiplicative group of integers $\pmod {17}$? How would the test work? Thanks for help.