So first off, in my algebra course we've so far covered integers, rings, the Chinese remainder theorem, ideals, fields (including quadratics over $\mathbb{Z}_{p}$ ) and polynomials (definitions, the ring of polynomials, evaluating polynomials and division). We have not covered irreducible polynomials – in researching this I can only seem to find people working out the number of irreducible monic polynomials.
Honestly, I've not got far with this question, I'm really unsure how I am supposed to work this out. All I really have is that I'm considering how many elements are in the set $\{X^2+bX+c : b,c \in \mathbb{Z}_{23}\}$ which I would assume is $23^2 = 529$ as we have $23$ choices for $b$ and $23$ coices for $c$. Is this question that simple, and the answer is just $529$? Or am I missing something?
Then for the second part, I know that all of the squares in $\mathbb{Z}_{23}$ are $S = \{0, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18\}$ and so for any of these monic polynomials to have a solution in $\mathbb{Z}_{23}$ we must have
$$
\Delta = b^2 – 4c \in S
$$
However, I'm really struggling to simplify this and work out how many such polynomials there are, without going through a laborious brute force approach.
Best Answer
For part 1, yes, it really is that simple. There are $23^2 = 529$ distinct monic quadratic polynomials.
For part 2, I would suggest you look at not the coefficients and the discriminant, but at the roots themselves. If the quadratic has roots in $\Bbb Z_{23}$, then either it's a double root ($\binom{23}1 = 23$ possible choices), or it's two distinct roots ($\binom{23}2 = 253$ possibilities). Since a monic polynomial is uniquely determined by its roots and their multiplicities, this does indeed count the polynomials you're after.
To be honest, this is how I would personally go about counting the number of irreducible monic quadratics anyways (that number is $529 - 23 - 253$). It wouldn't surprise me if some of the sources you found in your research actually did it this way too.
Addendum to address the concerns in the comment section:
If a quadratic has a double root, then that doube root is the only root. There are $23$ elements in $\Bbb Z_{23}$, and each of them can be that double root. The resulting polynomials are $$ (x-0)^2, (x-1)^2, (x-2)^2, \ldots, (x-22)^2 $$ As for the correspondence between monic polynomials and their roots, I am pretty sure you have seen that fact at some point, maybe even in high school, but we can do a quick proof here too.
Clearly linear monic polynomials are uniquely determined by their roots (a monic linear polynomial is of the form $x-a$, and its root is $a$, and if a monic linear polynomial has root $a$, then it must be equal to $x-a$).
Now we look at quadratic polynomials, say we have two monic quadratic polynomials, $f$ and $g$, both with the same roots $s, t$. Which is to say, $f(s) = g(s) = f(t) = g(t) = 0$, and no other $f(r)$ or $g(r)$ is zero. (Note that it is possible that $s = t$, and that case should be looked at separately. Here we will assume $s\neq t$ for simplicity.)
By polynomial division of $f$ and $g$ by $x-s$, we have $$ f(x) = (x-s)f_0(x) + a\\ g(x) = (x-s)g_0(x) + b $$ where $f_0, g_0$ are monic linear polynomials and $a, b$ are constants. Inserting $x = s$ into the above, we see that $f(s) = 0$ and $(s-s)f_0(s) = 0$, so we must have $a = 0$. Similarly $b = 0$. Inserting $x = t$ instead now tells us that $f(t) = 0$ and $t-s\neq 0$, so we must have $f_0(t) = 0$. Thus $f_0(x) = x-t$. Similarly $g_0 = x-t$ as well. So we therefore have $$ f(x) = (x-s)(x-t) = g(x) $$ and therefore two monic quadratic polynomials with the same roots are necessarily equal. (Note that this exact procedure won't hold when $s = t$, but we can show that in that case we must necessarily have $f(x) = g(x) = (x-s)^2$).
Therefore, to count the number of polynomials with roots is the same as counting the possible combinations of roots. The proof for a general degree rather than just quadratic polynomials is very similar.