# Finding three numbers that are pairwise not relatively prime, but with $\gcd(a,b,c)=1$

Find integers $a,b,$ and $c$ where $\gcd(a,b,c) = 1$, but $\gcd(a,b) \neq 1$, $\gcd(a,c) \neq 1$, and $\gcd(b,c) \neq 1$.

I tried so many combinations but I can't find 3 integers that meet these requirements.

I even though $(0,0,0)$ works, because I tried to convince myself 1 is the first positive integer where 0 has a divisor, because you can't divide by 0. I am not sure if there is a more systematic approach to this.

• Hint: Try to imagine venn diagrams of prime factors Oct 25 '16 at 4:14
• Thank you. I didn't really want an answer, I wanted a way to try and solve it myself. Oct 25 '16 at 4:16
• No problem. Next time you can explicitly mention hints only Oct 25 '16 at 4:17

Consider $a = 6, b= 10, c = 15$
An easy way to construct these is by considering three prime $2, 3, 5$, then pairwise multiply them.
• As an exercise try to construct $a, b, c, d$ such that $gcd(a, b, c, d) =1$, but pair-wise and triple-wise has gcd greater than 1. =) Oct 25 '16 at 4:20
Choose any three prime numbers $p,q,r$ and create the products $pq,qr,pr$. Each product lacks one of the prime factors, so the largest number that divides all three is $1$, but every pair of products has a factor in common, and hence a gcd greater than $1$. This is a generalization of the rather specific answer given first.