Euler's totient function is the function $$$\phi(n) =$$$ the number of numbers $$$\le n$$$ and coprime to $$$n$$$. Sum that for all $$$y$$$ from $$$2$$$ to $$$n$$$

Hint: DP.

Solution: Let us solve the more general problem: how do you find the number of pairs ($$$p[d]$$$) of integers $$$(x, y)$$$ with a gcd of $$$d$$$?

Let $$$cnt[x]$$$ mean the number of times $$$x$$$ occurs as a divisor of a number in $$$a$$$. We would calculate this by iterating over all $$$a_i$$$ and finding its divisors, incrementing $$$cnt[d]$$$ for each divisor (in $$$O(n \sqrt[3]{n}$$$)

What if $$$d=\max{a_i}$$$? Then the answer is $$$\frac{cnt[d] \cdot (cnt[d]-1)}{2}$$$. Otherwise, let us initialize the answer for some $$$x<\max{a_i}$$$ to $$$\frac{cnt[x] \cdot (cnt[x]-1)}{2}$$$. This is almost correct, but will also count pairs with a gcd of $$$2x$$$, $$$3x$$$, $$$4x$$$, and so on. So we will subtract $$$p[2x]$$$, $$$p[3x]$$$, $$$p[4x]$$$. This will take $$$O(n \log{n})$$$ time.

Here's my code for Counting Coprime Pairs on CSES. Hope this helped!

It can be done fast using mobius inversion: https://mirror.codeforces.com/blog/entry/53925