More on $\phi$ in 2016

The function is multiplicative, therefore,

$\begin{aligned} \sum_{d|2016^2} \phi (d) & = \sum_{d|2^{10}3^47^2} \phi (d) \\ & = \sum_{d|2^{10}} \phi (d) \sum_{d|3^4} \phi (d) \sum_{d|7^2} \phi (d) \\ & = (1+1+2+4+8+16+32+64 +128+256+512)(1+2+6+18+54)(1+6+42) \\ & = 1024 \times 81 \times 49 = \boxed{4064256} \end{aligned}$

We observe that the function $f(n)=\sum_{d|n}\phi(d)$ is multiplicative, being the Dirichlet convolution of $\phi$ and the constant function 1. Now we see that $f(p^k)$ $=1+(p-1)+(p^2-p)+...+(p^k-p^{k-1})=p^k$ for primes $p$ , so that $f(n)=\sum_{d|n}\phi(d)=n$ for all $n$ . Thus $f(2016^2)=2016^2=\boxed{4064256}$

Just for now

$\large {\sum_{d \mid y}\phi(d) = y}$

There's an algebraic method, but you can solve this strictly by thinking about what $\phi(x)$ represents. (Hint: $\phi(x)$ represents the number of integers $<x$ which are coprime with $x$ )