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