Question

Prove or disprove: If gcd(a,b) = 1, then gcd (a+b,ab)=1

Answer 1

@KingGeorge

Answer 2

I'm pretty sure this is not true, so we just have to find a counterexample.

Answer 3

On second thought, this may actually be true. So let's try to prove it.

Answer 4

Hmm..what do you think of this?

Suppose gcd (a+b,ab) = d, then d|a+b and d|ab. Since d|ab, then either d|a or d|b. However, d|a+b so d|a AND a|b. Since d|a and d|b, then d=gcd(a,b). Contradiction.

Answer 5

Your argument would work if you assumed \(d\) were prime. This proof gets around that by just taking an arbitrary prime factor.

Suppose that \(\gcd(a+b,ab)=d\), and \(\gcd(a,b)=1\). Then let \(p\) be a prime number dividing \(d\). Then \(p|a+b\) and \(p|ab\). In particular, \(p\) must divide either \(a\) or \(b\). But since \(\gcd(a,b)=1\), \(p\) can't divide both \(a\) and \(b\). But now, since \(p\) is prime, that means \(p\nmid a+b\). This shows that \(\gcd(a+b,ab)=1\), which is what we wanted.

Answer 6

So the statement is true then?

Answer 7

It is true. Against my initial intuition.

Answer 8

Okay, this one is a tricky one!!