I would like to ask for some help about the following problem.
Given is a polynomial $\,f(x)=ax^{2}+bx+c\,$ in $\,\mathbb{Z}/ n\mathbb{Z},\,$ we know that this quadratic equation $\,f(x)=0\,$ has exactly 8 solutions. Find the smallest possible $n$ and the coefficients $a$, $b$ and $c$.
Clearly, here we have to deal with the Chinese Remainder Theorem. I guess we have to find integers $\,m_{1}$, $m_{2}$, $m_{3}$, $m_{4}$, such that $\mathbb{Z}_{n}\simeq \mathbb{Z}_{m_{1}}\oplus \mathbb{Z}_{m_{2}}\oplus \mathbb{Z}_{m_{3}}\oplus \mathbb{Z}_{m_{4}}$, so $n= m_{1}m_{2}m_{3}m_{4}$, because to have 8 solutions, we need that every $\mathbb{Z}_{m_{i}}$ delivers two solutions (this means the determinant must be strictly bigger than $0$).
The problem is, i don't know how to find all this numbers. Can anyone help me with this problem, please? Is my idea in general correct?
Thank you in advance!