$a,b,c$ 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 $c \not = 0$.

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 $3$.
