# 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!

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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 $mn$?
As I understood, if: $p(x) \equiv 0 \text{ (mod } m)\\$ $p(x) \equiv 0 \text{ (mod } n)\\$ Has a solution, which is also unique modulo mn. This solution has the form: $\\ p(x) \equiv [0 * n * (n^{-1} \mod q) + 0 * m * (m^{-1} \mod p)] \mod mn\\$ With: $n^{-1}=inv(n,m); \ m^{-1}=inv(m,n)$....I think this should be :) –  darkmeow Jul 9 '14 at 1:29