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    #21 Thread Author
    Quote Originally Posted by I like Serena View Post
    Starting somewhere, what happened to $b_1$ and $b_{\phi(m)}$?
    $b_1$ is always $1$ and $b_{\phi(m)}$ is always equal to $m-1$.

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    #22
    Quote Originally Posted by evinda View Post
    $b_1$ is always $1$ and $b_{\phi(m)}$ is always equal to $m-1$.
    Ah, okay. Perhaps it'd be useful to mention that.
    In particular it means that whatever $m$ is, 1 and -1 will always be their own inverse, and they will from a pair that are additive inverses.
    (Assuming $m>2$.)

    Otherwise, it seems your proof is correct.

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    #23 Thread Author
    Quote Originally Posted by I like Serena View Post
    Ah, okay. Perhaps it'd be useful to mention that.
    Yes, I agree...


    Quote Originally Posted by I like Serena View Post
    In particular it means that whatever $m$ is, 1 and -1 will always be their own inverse, and they will from a pair that are additive inverses.
    (Assuming $m>2$.)


    Quote Originally Posted by I like Serena View Post


    Otherwise, it seems your proof is correct.
    Nice... Thank you!!!

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