1
$\begingroup$

$ 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!

$\endgroup$
2
$\begingroup$

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 $mn$?

$\endgroup$
1
  • $\begingroup$ 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 :) $\endgroup$ – darkmeow Jul 9 '14 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.