3. A function $g: \mathbb{N} \rightarrow \mathbb{N}$ satisfies the following:
(a) If $m$ is a proper divisor of $n$, then $g(m)<g(n)$.
(b) If $m$ and $n$ are relatively prime and greater than 1 , then
$$
g(m n)=g(m) g(n)+(n+1) g(m)+(m+1) g(n)+m+n \text {. }
$$

Find the least possible value of $g(2016)$.
3. Consider $h(x):=g(x)+x+1$. We have that, for $m, n$ coprime and greater than 1 ,
$$
\begin{aligned}
h(m) h(n) & =(g(m)+m+1)(g(n)+n+1) \\
& =g(m) g(n)+(n+1) g(m)+(m+1) g(n)+m n+m+n+1 \\
& =g(m n)+m n+1 \\
& =h(m n) .
\end{aligned}
$$

Repeating this, we find that more generally, if $m_{1}, m_{2}, \ldots, m_{k}$ are pairwise coprime positive integers all greater than 1 ,
$$
h\left(\prod_{i=1}^{k} m_{i}\right)=\prod_{i=1}^{k} m_{i} .
$$

Hence, it suffices to consider $h$, and thus $g$, only on prime powers. Since
$$
g\left(p^{n}\right)>g\left(p^{n-1}\right)>\cdots>g(p)>g(1) \geq 1,
$$
we have $g\left(p^{n}\right) \geq n+1$. Indeed, taking $g(1)=1, g\left(p^{n}\right)=n+1$ gives us a well-defined function $g$ on $\mathbb{N}$. To solve for $g(2016)$, we solve for $h(2016)$ first, noting that $2016=2^{5} \cdot 3^{2} \cdot 7^{1}$ :
$$
\begin{aligned}
h(2016) & =h\left(2^{5}\right) h\left(3^{2}\right) h\left(7^{1}\right) \\
& =\left(7+2^{5}\right)\left(4+3^{2}\right)\left(3+7^{1}\right) \\
& =5070
\end{aligned}
$$
and so $g(2016)=5070-2017=3053$. This is the minimum possible value of $g(2016)$.