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Math Help - relatively prime

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    relatively prime

    How do I prove that two consecutive integers are relatively prime? In other words, how do I show if two integers have no common factors other than 1? Thanks
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    Quote Originally Posted by TitaniumX View Post
    How do I prove that two consecutive integers are relatively prime? In other words, how do I show if two integers have no common factors other than 1? Thanks
    Let n and n+1 be the 2 consecutive integers.

    Let d be a common divisor of n and n+1.
    Hence n=dk' and n+1=dk'' where k' and k'' are integers.

    So (n+1)-n=1=d(k'-k'')

    ------> d must divide 1. which is not possible unless d=1
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    Suppose you have some integer  N_0 whose factorization is
     p_0^{a_0}p_1^{a_1}...p_t^{a_t} . The consecutive integer is
     N_1 = p_0^{a_0}p_1^{a_1}...p_t^{a_t} + 1 . Let  P be a prime dividing  N_1 . Then  P doesn't divide  N_0 otherwise it would divide the difference  N_1-N_0 = 1 . This is impossible.
    Therefore, there is no prime integer that divide both  N_0 and  N_1 . In other words they have no common factors or are relatively prime.
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