a/ Solve $$\large f(m + n) = f(m) + f(n) + mn$$ for $f \colon \mathbb N \to \mathbb N$.
b/ Solve $$\large g(mn) = g(m)g(n)(m + n)$$ for $f \colon \mathbb N \to \mathbb N$.
I'm only getting started on solving function equations so if there is anything I don't get correctly.
This is my attempt at this problem.
For a/, replacing $m$ with $m - 1, m - 2, \cdots, 2, 1$ respectively, we have that
$$f(m) - f(m - 1) = f(1) + (m - 1)$$
$$f(m - 1) - f(m - 2) = f(1) + (m - 2)$$
$$\vdots$$
$$f(3) - f(2) = f(1) + 2$$
$$f(2) - f(1) = f(1) + 1$$
$$\implies f(m) - f(1) = (m - 1)f(1) + \frac{m(m - 1)}{2}$$
$$\implies f(m) = m\left[f(1) + \frac{m - 1}{2}\right]$$
Therefore, $f(x) = x\left[f(1) + \dfrac{x - 1}{2}\right], \forall x \in \mathbb N, x \ne 1$ and $f(1) = a, a \in \mathbb R$.
For b/, we have that $g(m) = g(m)g(1)(m + 1)$ where $n = 1$.
$\iff g(1) = \dfrac{1}{m + 1} \implies m$ is fixed, which is false.
So there aren't any such functions $g(x)$ such that $g(mn) = g(m)g(n)(m + n)$ over $f \colon \mathbb N \to \mathbb N$.
First equation:
If we write $g(n)= 2f(n)-n^2$ then we have to solve Cauchy equation $$g(m+n)=g(m)+g(n)$$ for $g:\mathbb{Z}\mapsto \mathbb{Z}$ which is $g(n) = cn$ for some integer $c$.
So $$f(n)= {cn+n^2\over 2}$$ and clearly $c$ must be odd and positive.