Sure! So we are gonna define $$$dp[v]$$$ as the number of pairs in the array that have a gcd of $$$v$$$. How many pairs could potentially have a gcd of $$$v$$$? That would be all the multiples of $$$v$$$. Of course, for any $$$a,\,b$$$ where $$$a$$$ and $$$b$$$ are both multiples of $$$v$$$, we could have the case where $$$gcd(a, b) \ne v$$$. However, notice that $$$gcd(a, b)$$$ must be a multiple of $$$v$$$. Therefore, if we have $$$dp[m \cdot v]$$$ already calculated, where $$$m \ge 2$$$, we can calculate $$$dp[v]$$$ as $$$\frac{n \cdot (n - 1)}{2} - \sum_{m = 2}^{\infty} dp[m \cdot v]$$$, where $$$n$$$ is the number of elements in the array that are multiples of $$$v$$$. This relationship basically says that $$$dp[v]$$$ is the number of pairs of multiples of $$$v$$$ minus the number of pairs of multiples of $$$v$$$ that have a gcd that is a multiple of $$$v$$$ (and not equal to $$$v$$$). You can see that, upon doing this, all we are left with are the pairs of multiples of $$$v$$$ that have a gcd of $$$v$$$.

Of course, you don't have to do the summation ad infinitum. Because there will never be a pair $$$(a,\,b)$$$ in the array with $$$gcd(a, b) > A$$$, where $$$A$$$ is the maximum element in the array, we only have to create a $$$dp$$$ array that stores values for elements $$$1..A$$$ inclusive.

The final answer will be $$$\frac{N \cdot (N - 1)}{2} - dp[1]$$$ — the number of pairs in the entire array with a $$$gcd \ne 1$$$. I hope that this has helped.

EDIT: I just realized that I misread the problem. In the case of two arrays, you can create an array $$$C = A + B$$$ and run this $$$dp$$$ on $$$C$$$. Then you can run this $$$dp$$$ on each individual array $$$A$$$ and $$$B$$$ and the final answer should be $$$answer_C - answer_A - answer_B$$$.