Quadratic Polynomials in $Z$ mod 3

Question

In my number theory class my professor wrote that

A quadratic polynomial over $\mathbb{Z}_3$ has the form $ax^2+bx+c$, where $a \ne 0$

Then he wrote that “There are 18 Quadratic polynomials over $Z_3$”

My question is how did he get $18$?

If I have
$2$ choices for $a$
$3$ choices for $b$
$2$ choices for $c$

Then how can I have $\mathbf{18}$ Quadratic polynomials over $\mathbb{Z}_3$?

Answer 1

You have $2$ choices for $a$, $3$ choices for $b$, and $3$ choices for $c$. That gives you indeed $2\times3\times3=18$ possibilities.

Answer 2

They are of the form $ax^2 + bx +c$, where $a \neq 0$.

Since we are in $\mathbb{Z}_3$, the possible values for $b$ and $c$ are ${0,1,2}$, and the possible values for $a$ are 1,2 (since $a \neq 0)$. Now we count all the different possible combinations, so the number of different quadratic polynomials over $\mathbb{Z}_3$ is given by $3\cdot 3 \cdot 2 = 18$.

Answer 3

There are three choices for $c$, not two. For example, the following are all different module $3$:

$$x^2+x\quad x^2+x+1\quad x^2+x+2$$

To prove that these are all different, notice that they all evaluate to different values at $x=0$. Alternatively, you should be able to easily prove they all fall into different equivalence classes by contradiction.