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Thread: Greatest common divisor

  1. #1
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    Greatest common divisor

    Hi.
    Can anybody prove that?

    Let $\displaystyle n>4$

    $\displaystyle GCD(n, (n-1)!)=1 \Longleftrightarrow{} n$ its prime.

    and

    if $\displaystyle n$ its not prime $\displaystyle \Longrightarrow{} GCD(n, (n-1)!)=n$

    Thanks
    Last edited by asub1; Dec 3rd 2008 at 01:02 PM.
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  2. #2
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    Quote Originally Posted by asub1 View Post
    $\displaystyle GCD(n, (n-1)!)=1 \Longleftrightarrow{} n$ its prime.
    This should be clear. If $\displaystyle n$ is prime then $\displaystyle 1,2,3...,n-1$ show no non-trivial factors with $\displaystyle n$. Thus, $\displaystyle (n,(n-1)!)=1$.

    if $\displaystyle n$ its not prime $\displaystyle \Longrightarrow{} GCD(n, (n-1)!)=n$
    If $\displaystyle n$ is not prime then $\displaystyle n=ab$ where $\displaystyle 1<a,b<n$. Therefore, we are able to find $\displaystyle a,b$ among $\displaystyle 1,2,...,n-1$.
    Thus, $\displaystyle n|(n-1)!$ and so $\displaystyle (n,(n-1)!)=n$.
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  3. #3
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    Hi!


    If n is not prime then $\displaystyle n=ab$ where $\displaystyle 1< a,b <n$. Therefore, we are able to find $\displaystyle a, b$ among $\displaystyle 1, 2, ..., n-1$.
    Thus, $\displaystyle n|(n-1)!$ and so $\displaystyle (n, (n-1)!)=n$.
    What about $\displaystyle a=b$?
    $\displaystyle n=aa$ where $\displaystyle 1<a, a<n$. we are able to find $\displaystyle a$ among $\displaystyle 1, 2, ..., n-1$, but... and the other $\displaystyle a$?

    Can you prove?

    Thanks
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