## Monday Math 149

What is the probability that two independently randomly chosen integers are mutually prime (have no common factor greater than 1)? The probability for four random integers? For n integers in general?

The probability that a random integer is divisible by a given prime p is $\frac1p$ (since every pth integer is a multiple of p); thus, the probability of two independent random integers are both divisible by p is $\frac1{p^2}$. Thus, the probability that they do not have p as a common factor is $1-\frac1{p^2}$.
Thus, for the two random integers to be mutually prime, they must have no primes as a common factor; thus, the probability is the product of the above for all primes,
$\prod\limits_{p\;\text{prime}}\left(1-p^{-2}\right)=\left(1-2^{-2}\right)\left(1-3^{-2}\right)\left(1-5^{-2}\right)\cdots$
Now, recalling from here that $\zeta(s)=\prod\limits_{p\;\text{prime}}\frac1{1-p^{-s}}$, thus the probability is
$\prod\limits_{p\;\text{prime}}\left(1-p^{-2}\right)=\frac1{\zeta(2)}$,
and since we found here that $\zeta(2)=\frac{\pi^2}{6}$, we see that the probability that two random integers are mutually prime is $\frac{6}{\pi^2}$, which is approximately 60.8%.
Similarly, the probability of four integers being all divisible by a prime p is $\frac1{p^4}$, so the probability that they do not all have p as a common factor is $1-\frac1{p^4}$. Analogous to the above, the probability that the greatest common factor of all four integers is one is
$\prod\limits_{p\;\text{prime}}\left(1-p^{-4}\right)=\frac1{\zeta(4)}$,
and since $\zeta(4)=\frac{\pi^4}{90}$ (see here); we see that the probability for four integers is $\frac{90}{\pi^4}$, which is approximately 92.4%.
Generalizing, we see that for n≥2 independent random integers, the probability that the greatest common factor of all n integers is 1 is given by $\frac1{\zeta(n)}$, which increases toward unity as n gets larger
(since $\zeta(s)=\sum\limits_{n=1}^{\infty}\frac1{n^s}$ decreases asymptotically toward 1 for increasing real s>1).