# Thread:

1. ## 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.

2. Originally Posted by 1337h4x
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.
$a = 2\times 3\times 5,$
$b = 2\times 3\times 7,$
$c = 2\times 5\times 7,$
$d = 3\times 5\times 7.$

3. 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.

4. Originally Posted by 1337h4x
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 n–1 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 $p$s except for $p_k$. Then $p_k$ will divide all the $a$s except for $a_k$. But there will be no common divisor (greater than 1) of all the $a$s.
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