# Relatively Prime Set of Integers

#### 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.

#### Opalg

MHF Hall of Honor
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.
$$\displaystyle a = 2\times 3\times 5,$$
$$\displaystyle b = 2\times 3\times 7,$$
$$\displaystyle c = 2\times 5\times 7,$$
$$\displaystyle d = 3\times 5\times 7.$$

#### 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.

#### Opalg

MHF Hall of Honor
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.

• Bruno J.