31.5 The Chinese remainder theorem
31.5-1
Find all solutions to the equations \(x \equiv 4 \pmod 5\) and \(x \equiv 5 \pmod{11}\).
\[
\begin{aligned}
m_1 & = 11, m_2 = 5. \\\\
m_1^{-1} & = 1, m_2^{-1} = 9. \\\\
c_1 & = 11, c_2 = 45. \\\\
a & = (c_1 \cdot a_1 + c_2 \cdot a_2) \mod (n_1 \cdot n_2) \\\\
& = (11 \cdot 4 + 45 \cdot 5) \mod 55 = 49.
\end{aligned}
\]
31.5-2
Find all integers \(x\) that leave remainders \(1\), \(2\), \(3\) when divided by \(9\), \(8\), \(7\) respectively.
\(10 + 504i\), \(i \in \mathbb Z\).
31.5-3
Argue that, under the definitions of Theorem 31.27, if \(\gcd(a, n) = 1\), then
\[(a^{-1} \mod n) \leftrightarrow ((a_1^{-1} \mod n_1), (a_2^{-1} \mod n_2), \ldots, (a_k^{-1} \mod n_k)).\]
\[\gcd(a, n) = 1 \rightarrow \gcd(a, n_i) = 1.\]
31.5-4
Under the definitions of Theorem 31.27, prove that for any polynomial \(f\), the number of roots of the equation \(f(x) \equiv 0 (\mod n)\) equals the product of the number of roots of each of the equations
\[f(x) \equiv 0 \pmod{n_1}, f(x) \equiv 0 \pmod{n_2}, \ldots, f(x) \equiv 0 \pmod{n_k}.\]
Based on \(\text{31.28}\)–\(\text{31.30}\).
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