It's a bit hazy

How do we solve this? It's just to notice that you can just "slide" each $n$ th row a bit to the right (or left, if you like) proportional to $n$ to obtain something resembling the Gaussian integers. The Gaussian case is well-known, but here a quick proof:

The probability that two integers share a certain prime factor $p$ is $\frac{1}{p} \times \frac{1}{p} = p^{-2}$ . The probability they are coprime (equivalent to visibilty) is the complement, ie $1-p^{-2}$ .

Take the product over all primes $p$ and you get the Euler product

$\prod_{p \ \text{prime}} \left( 1-p^{-2} \right) = \frac{1}{\zeta(2)} \approx \boxed{0.608}$

Two distinct primes p and q are always relatively prime, (p,q)=1, as are any positive integer powers of distinct primes p and q, (p^m,q^n)=1.

Relative primality is not transitive. For example, (2,3)=1 and (3,4)=1, but (2,4)=2.

The probability that two integers m and n picked at random are relatively prime is

P((m,n)=1)=[zeta(2)]^(-1)=6/(pi^2)=0.60792...

Jake Lai, it's probably a good idea to include a short definition of an Eisenstein integer in your problem statement.

The probability that two random numbers are pairwise coprime is $\frac{6}{\pi^2}$ .This can be shown by using the fact that the partial sums of $\phi(n)$ are $\frac{3}{\pi^2}x^2+O(xlogx)$ .By the way @JakeLai , from what book are you getting these problems from?