$a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $3$) holds for all $x$? Today I wrote a computer program that finds whole positive $a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $n$) holds for all $x$. For $n=2$ I get several results, such as $x^2+x$, $3x^2+x$ and more. However, I was not able to find $a,b,c$ for the case $n=3$.

(Why) Are there no $a,b,c$ for $n=3$?

Remark: Of course $\text{gcd}(a,b,c) = 1$ and $a,b,c \in \mathbb{Z}$ and $a \not = 0$ or $b \not = 0$ or $c \not = 0$.
 A: I assume $a,b,c$ and $x$ are all restricted to integers.
Considering the three cases $x = 0,1,2 \mod 3$ we have:
$x=0 \mod 3 \Rightarrow c=0 \mod 3$
$x=1 \mod 3 \Rightarrow a+b+c=0 \mod 3$
$x=2 \mod 3 \Rightarrow a +2b+c = 0 \mod 3$
These three simultaneous equations are only all satisfied if $a=b=c=0 \mod 3$ i.e. if $a$, $b$ and $c$ are all multiples of $3$.
A: What conditions are needed in order to conclude that the familiar quadratic formula holds, saying that if $ax^2+bx+c=0$ and $a\ne0,$ then $x= \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\text{ ?}$
One needs $2a$ to have a multiplicative inverse, regardless of which non-zero element $a$ is. Thus if $n=6,$ for example, and $a=3,$ you get $2a=0$ and that has no multiplicative inverse; thus dividing by $2a$ makes no sense.
But when $n=3$ and $a\ne 0$, then either $a=1$ and $2a=2,$ and $2$ is its own multiplicative inverse, or $a=2$ and $2a=1,$ and $1$ is its own multiplicative inverse. Thus the quadratic formula holds, so a quadratic equation cannot have more than two roots.
A: For any odd n > 2 the only solution is a=b=c=0.
x=0 gives c=0.
x=1 gives a+b=0; x=-1 gives a-b=0. These give 2a=2b=0.
If n is odd, these give a=b=0.
If n is even, there can be solutions such as
$(n/2)x^2+(n/2)x$.
