Set of Solutions of A Quadratic Equation with Coefficients in $\{0,1,\cdots , \ p-1\}$

Question

I was just playing with quadratic equations and this interesting question came into my mind. Say I have a set of quadratic polynomials $S=\{f_{(b,c)}(x)=x^2+bx+c:b,c\in \{0,1,\cdots, p-1 \}\}$ where $p\in \mathbb{N},\ p\ge 1$. I want to investigate how do the roots of the equations $f_{(b,c)}(x)=0$ vary over the complex plane as $b,c$ are varied. I found all the roots for the equations $f_{(b,c)}=0$ for the case $p=2$, i.e. when $b,c\in\{0,1\}$. I found the roots as $-1,\pm i,\omega,\omega^2,0,0,0$. So $3$ roots are at the origin and the rest of $5$ are on the unit circle $|z|=1$. I found this structure very intriguing. So, is there any general result on this?

Answer 1

Consider two cases:

1. The roots are real
   The vertex of the parabola is at coordinates $$V\left( -\frac b2, c-\frac{b^2}4 \right)$$ The roots can not be greater than $0$, since the coefficients of the polynomial are not negative. Moreover, the parabola has always the same shape, since the coefficint of $x^2$ is $1$. The modulus of the roots will be maximum when the vertex is most below and most left (sorry for my English), that is, when $b=p-1$ and $c=0$. This gives the root $1-p$.

2. The roots are not real
   Now, the real part is $-b/2$ and the imaginary part is $$\pm\frac{\sqrt{4c-b^2}}2$$

Regardless of sign, the modulus is $$\frac{b^2}4+\frac{4c-b^2}4=c$$ so, again, the moduls of the roots is at most $p-1$.

Summing up, the roots of the polynomials are always on the disc of radius $p-1$. By the way, note that I have made no use for the fact that $p$ is prime, or even integer.