Multiplicative functions and Chinese remainder theorem $ p $ is a nonconstant polynomial with integer coefficients.Define the function $\chi_p(n)$ as the number of zeros of $ p $ in $\mathbb{Z}_n$ for $ n > 1 $, and $ \chi_p(1) = 1 $. e.g., consider $ p(x) = x^2 + 1 $, see table (Zeros of $p(x) = x^2 + 1 \mod n$)
$$ 
\begin{align}
&n && Zeros && \chi_p(n)\\
&===&&=== &&===\\
&2 && \{ 1 \}       && 1 \\
&3 && \varnothing       && 0 \\
&4 && \varnothing       && 0 \\
&5 && \{ 2, 3 \}       && 2 \\
&6 && \varnothing       && 0 \\
&10 && \{ 3, 7 \}       && 2 \\
&13 && \{ 5, 8 \}       && 2 \\
&15 && \varnothing       && 0 \\
&65 && \{ 8, 18, 47, 57 \}  && 4 \\
\end{align} $$
Prove $ \chi_p $ is multiplicative, considering the zeros of $ p \mod mn $, if $ m $ and $ n $ are relatively prime, and applying the Chinese remainder theorem.
Help Please!I do not know how to start!

Hint: By the Chinese Remainder Theorem, $$p(x) \equiv 0 \text{ (mod } mn) \iff p(x) \equiv 0 \text{ (mod } m) \text{ and }p(x) \equiv 0\text{ (mod } n)$$
If $p(x) \equiv 0 \text{ (mod } m)$ has $k$ solutions and $p(x) \equiv 0 \text{ (mod } n) $  has $j$ solutions, how many solutions are there modulo $mn$?
