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

• The question is unclear. $x^2 + x$ does not have $c$ positive. On the other hand, for any $n$ taking $a = b = c = n$ always gives a solution (in particular for $n = 3$), no? – Travis Willse Nov 24 '17 at 14:27