Relatively Prime Set of Integers

Apr 2009
96
0
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.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
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.
\(\displaystyle a = 2\times 3\times 5,\)
\(\displaystyle b = 2\times 3\times 7,\)
\(\displaystyle c = 2\times 5\times 7,\)
\(\displaystyle d = 3\times 5\times 7.\)
 
Apr 2009
96
0
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.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
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 \(\displaystyle a_1,a_2,\ldots,a_n\) 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 \(\displaystyle p_1,p_2,\ldots,p_n\). For \(\displaystyle 1\leqslant k\leqslant n\), let \(\displaystyle a_k\) be the product of all the \(\displaystyle p\)s except for \(\displaystyle p_k\). Then \(\displaystyle p_k\) will divide all the \(\displaystyle a\)s except for \(\displaystyle a_k\). But there will be no common divisor (greater than 1) of all the \(\displaystyle a\)s.
 
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