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Math Help - x^2\equiv 1 (mod p) and x^2\equiv 1 (mod q), where p and q are distinct primes

  1. #1
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    x^2\equiv 1 (mod p) and x^2\equiv 1 (mod q), where p and q are distinct primes

    x^2\equiv 1 \ \mbox{(mod p) and} x^2\equiv 1 \ \mbox{(mod q)}, where p and q are distinct primes, it follows x^2\equiv 1 \ \mbox{(mod pq)?}

    p|(x^2-1) \ \mbox{and} \ q|(x^2-1)

    pq|(x^2-1)^2\rightarrow (x^2-1)^2\equiv 0 \ \mbox{(mod pq)}

    \rightarrow (x^2-1)^2\equiv 0 \ \mbox{(mod pq)}\rightarrow 1^2\equiv 0 \ \mbox{(mod pq)}\rightarrow 1\not\equiv 0 \ \mbox{(mod pq)}.

    Since this isn't true, then no??
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by dwsmith View Post
    x^2\equiv 1 \ \mbox{(mod p) and} x^2\equiv 1 \ \mbox{(mod q)}, where p and q are distinct primes, it follows x^2\equiv 1 \ \mbox{(mod pq)?}
    This is true.

    In general a \equiv b\ (\text{mod}\ m) and a \equiv  b\ (\text{mod}\ n) implies a \equiv b\ (\text{mod}\  \text{lcm}(m,n)). See this MathWorld page, property #11.

    Quote Originally Posted by dwsmith View Post
    (x^2-1)^2\equiv 0 \ \mbox{(mod pq)}\rightarrow 1^2\equiv 0 \ \mbox{(mod pq)}
    I don't see how you made this step.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    x^2\equiv 1 \ \mbox{(mod p) and} x^2\equiv 1 \ \mbox{(mod q)}, where p and q are distinct primes, it follows x^2\equiv 1 \ \mbox{(mod pq)?}
    This is correct and you have the right idea on how to prove it.

    See here for full a proof.
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