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Thread: Gcd problem

  1. #1
    Junior Member
    Joined
    May 2009
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    36

    Gcd problem

    Proof that:


    gcd (n!+1 , (n+1)! + 1) = 1
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  2. #2
    Junior Member
    Joined
    Aug 2009
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    Gothenburg, Sweden
    Posts
    37
    Suppose that there exists a $\displaystyle p \geq 2$ such that $\displaystyle p \mid n!+1$. If $\displaystyle p \mid a$ then $\displaystyle p \mid ca$ where c is an integer. Using this $\displaystyle p \mid (n+1)(n!+1)$. Now look at $\displaystyle (n+1)!+1$. If $\displaystyle p \mid (n+1)!+1=(n+1)(n!+1)-n$ we come to the conclussion that $\displaystyle p \mid -n$ but this is impossible since $\displaystyle n!+1 \equiv 1 \pmod n$ so p=1 and this is a contradiction so there exist no number p such that p divides $\displaystyle n!+1$ and $\displaystyle (n+1)!+1$ and it follows that $\displaystyle GCD(n!+1,(n+1)!+1)=1$
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