Use a range query data structure such as segment tree/sparse table.

The solution that I found most common after having solved the problem was using a range query data structure to answer queries on range [L, R] efficiently. I used a RQ DS too and so I'll explain a solution that uses segment tree.

For a fixed point L, find the left most right boundary R such that $$$gcd([L, ..., R])$$$ is 1. Once we have such a value R, we can now move the left boundary. For any index i in [L, R], the $$$gcd([i, ..., R])$$$ will either be 1 or something greater. When we encounter a left boundary that makes gcd to something not 1, we start searching for another right boundary keeping the left boundary fixed. Among all such [L, R] endpoints, the one with maximum length is your answer.

The solution using RQ DS is, imo, the most intuitive. The complexity you achieve this way is $$$O(n.log(n))$$$. You can use either segment tree or sparse table or any other logarithmic query associative DS to achieve this complexity. Note that the complexity actually seems like it's $$$O(n.log^2(n))$$$ because of gcd being in there but finding the gcd of an entire array is magically $$$O(n)$$$ approximately and not $$$O(n.({some \space log \space factor}))$$$.

There also is a solution that uses something called a gcdqueue which is similar in structure to this but I do not understand it and will not explain.