6

If we pick any two positive integers at random, what is the probability that their greatest common factor is 2? I have been wondering about this problem for a while and done some work on it. I started by thinking of a reasonable upper or lower bound. My upper bound was simply $.25$ as that is the probability of picking two even numbers (i.e. having a factor of 2) but if I was to pick (8,12) their gcf is 4. Then I thought about subtracting off the probability that the two numbers chosen had a common factor of 4 and then doing that for 6, 8, 10 and so on which got me to the sum of $1/4 -1/16 - 1/36 - 1/64 -...$ $= 1/2 - \frac{\pi^2}{24}$ I did some experimentation on smaller cases and I do not think my work was right, it should be closer to about $.158$ Also I think my method removed too much like when I subtracted off numbers that had 4 as a factor and numbers that had 6 as a factor I subtracted 12 twice. What should I do to adjust my work or get back on track? I was considering inclusion-exclusion but that seems too complicated for this problem. Thank you for any help

user498190
  • 151
  • 7
  • 1
    I think this is a great question, but it may not have a well-defined answer since there is no uniform probability distribution on the integers. In other words, you need some notion of how to pick a random integer to make the question precise. One possibility is to ask the question for the first N integers and then see if a limit exists as $N\to\infty$. – Cheerful Parsnip Jul 17 '19 at 03:47
  • That would make sense. The question does not explicitly say that each integer has equal probability. I sense that there will be a connection to the problem of "What is the probability that 2 integers are co-prime?" which is actually $\frac{6}{\pi^2}$ – user498190 Jul 17 '19 at 03:51

1 Answers1

6

In this question it is shown that the probability two integers are coprime is $\frac 6{\pi^2}$ in the sense that you choose the integers between $1$ and $N$ uniformly, then take the limit as $N \to \infty$. For the $\gcd$ to be $2$, you need both integers to be even, then the numbers that are half of each of them to be coprime. The chance is then $\frac 14 \cdot \frac 6{\pi^2}$

Ross Millikan
  • 383,099