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Math Help - Two consecutive numbers are relatively prime

  1. #1
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    Two consecutive numbers are relatively prime

    Two consecutive numbers are relatively prime.

    (n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1

    I am not sure what to do next to prove this.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    Two consecutive numbers are relatively prime.

    (n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1

    I am not sure what to do next to prove this.
     \alpha=-1,\; \beta=1
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  3. #3
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    I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?
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    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?
    Let  g=(n,n+1)

     g\mid n and  g\mid n+1 \implies g\mid (n+1)-n = 1 \implies g=\pm1 . But since  g>0 , we get  g=1 .
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?
    The answer is yes. In general: Let a and b be integers. There exists two integers x and y such that ax+by=1 if and only if (a,b)=1. In your case, a=n, x=-1 and b=n<br />
+1, y=1.

    However, the reply you got from "chiph588" using the greatest common divisor ( g=(n,n+1)) seems to me more elegant.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by melese View Post
    The answer is yes. In general: Let a and b be integers. There exists two integers x and y such that ax+by=1 if and only if (a,b)=1. In your case, a=n, x=-1 and b=n<br />
+1, y=1.
    This is a special case of Bézout's identity.
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