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I ran into this version of the Chinese Remainder Theorem:

Show that the solutions to the simultaneous system of congruences $$x\equiv a_1\pmod{m_1},$$ $$x\equiv a_2\pmod{m_2},$$ $$...$$ $$x\equiv a_r\pmod{m_r},$$ where the $m_j$ are pairwise relatively prime, are given by $$x\equiv a_1M_1^{\phi(m_1)}+a_2M_2^{\phi(m_2)}+\cdots+a_rM_r^{\phi(m_r)}\pmod{M},$$ where $M=m_1m_2\cdots m_r$ and $M_j=M/m_j$ for $j=1, 2, ..., r.$

This is basically the exact Chinese Remainder Theorem, except that instead of $x_jM_jy_j$, we have $x_jM_j^{\phi(m_j)}$. Would it then suffice to show that $x_jM_j^{\phi(m_j)}\equiv x_jM_jy_j\pmod{M}$? If so, how could I go about it? Thanks in advance!

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  • $\begingroup$ Relevant: en.wikipedia.org/wiki/Euler's_theorem $\endgroup$
    – user2468
    Mar 7, 2012 at 1:55
  • $\begingroup$ You introduce the notation $y_j$ without definition. But what are you trying to accomplish? Are you trying to prove the CRT in the given version? Isn't it enough to just take the proposed solution and show it satisfies the congruences, doing so by making use of Euler's Theorem? $\endgroup$
    – Gerry Myerson
    Mar 7, 2012 at 1:55
  • $\begingroup$ @GerryMyerson, $y_j$ is the inverse of $M_j$ modulus $m_j$. I will try what you suggested. $\endgroup$
    – wjmolina
    Mar 7, 2012 at 1:59

1 Answer 1

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Hint $\ $ Both $\rm\: M_i^{\phi(m_i)}$ and $\rm\:M_i ({M_i}^{-1}\ mod\ m_i)\:$ are $\rm\:\equiv 1\pmod{m_i},\:$ and $\rm\:\equiv 0\pmod{m_j},\ j\ne i$

since $\rm\:gcd(M_i,m_i) = 1,\:$ and $\rm\:j\ne i\:\Rightarrow\:m_j\ |\ M_i $

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