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Thread: primitive roots

  1. #1
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    primitive roots

    Let n>1 be a positive integer. Find the number of all m, so gcd(n,m)=1,
    $1\le m<n$ and $mn-1\equiv 1$ (mod n).
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  2. #2
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    Re: primitive roots

    Answer: If n = 2, then the answer is 1. If n>2, the answer is 0.

    This is because $mn-1 \equiv 1 \text{ (mod n)}$ if and only if $-1 \equiv 1 \text{ (mod n)}$. Note that $mn$ is a multiple of $n$ for any integers $m,n$, so that factor is equivalent to $0\text{ (mod n)}$.
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  3. #3
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    Re: primitive roots

    mn-1=1 \pmod{n}  \implies mn=2 \pmod{n}  \implies 0=2 \pmod{n}
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