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Thread: Relatively Prime Set of Integers

  1. #1
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    Relatively Prime Set of Integers

    Hello all,

    Does anyone know of a set of 4 integers S=[a,b,c,d] where where a,b,c,d >0, and that 3 of the integers in S have a common divisor 'x' > 1 , but also that the GCD(a,b,c,d)=1 ?

    This is essentially plugging and playing in my mind, but I'm sure there is a more sophisticated method to this madness.
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  2. #2
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    Quote Originally Posted by 1337h4x View Post
    Hello all,

    Does anyone know of a set of 4 integers S=[a,b,c,d] where where a,b,c,d >0, and that 3 of the integers in S have a common divisor 'x' > 1 , but also that the GCD(a,b,c,d)=1 ?

    This is essentially plugging and playing in my mind, but I'm sure there is a more sophisticated method to this madness.
    a = 2\times 3\times 5,
    b = 2\times 3\times 7,
    c = 2\times 5\times 7,
    d = 3\times 5\times 7.
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    How did you reach this conclusion? Is there some sort of generic way to solve this? For example, if there was a set of 5 integers where 4 had a common divisor > 1 but the GCD was still equal to 1.
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  4. #4
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    Quote Originally Posted by 1337h4x View Post
    How did you reach this conclusion? Is there some sort of generic way to solve this? For example, if there was a set of 5 integers where 4 had a common divisor > 1 but the GCD was still equal to 1.
    The pattern is this. Suppose you want to find a set of n positive integers a_1,a_2,\ldots,a_n for which every subset containing n1 of them has a common divisor > 1, but the GCD of the whole set of n integers is equal to 1. The method is to take n distinct prime numbers p_1,p_2,\ldots,p_n. For 1\leqslant k\leqslant n, let a_k be the product of all the ps except for p_k. Then p_k will divide all the as except for a_k. But there will be no common divisor (greater than 1) of all the as.
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