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$ 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 $mn$?

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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 at 1:29

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