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  1. #1
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    prove ....

    prove that if n>6 then  \exists a,b \in N \ a,b >1 , \ gcd (a,b)=1 such that n= a+b
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  2. #2
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    Quote Originally Posted by flower3 View Post
    prove that if n>6 then  \exists a,b \in N \ a,b >1 , \ gcd (a,b)=1 such that n= a+b

    If n is odd you can take a=n-2\,,\,\,b=2 . I'll let you the case for even n. Can you say now why they required n>6 ?

    Tonio
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  3. #3
    Super Member Bacterius's Avatar
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    Hello,
    n > 6. Consider two cases :

    \Rightarrow n is odd. Let a = \frac{n - 1}{2} and b = \frac{n + 1}{2}. a + b = \frac{2n}{2} = n, and b - a = \frac{n + 1 - n + 1}{2} = 1. Since b = a + 1, gcd(a, b) = 1.

    \Rightarrow n is even. Let a = \frac{n - \beta}{2} and b = \frac{n + \beta}{2}, for some \beta odd prime. a + b = \frac{2n}{2} = n, and b - a = \frac{n + \beta - n + \beta}{2} = \beta. Since b = a + \beta, gcd(a, b) = 1 if and only if gcd(a, \beta) = 1. Since a number cannot potentially divide every prime less than itself (otherwise it would at least be the product of all these primes and would be much greater, leaving more primes in between), there must exist some \beta for which gcd(a, \beta) = 1, and gcd(a, b) = 1 follows.

    As you can see, the case for n even is much harder than the case for n odd. Can you see how I used divisibility theorems and properties of prime numbers to solve this problem ? If you have any questions/concerns, feel free to ask.

    Last edited by Bacterius; February 24th 2010 at 04:04 AM. Reason: Typos
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