Question 54
Let $\mathbb{N}$ denote the set of all non-negative integers (including zero). A function $f: \mathbb{N} \to \mathbb{N}$ satisfies for all $a, b \in \mathbb{N}$
$$
f(a + b^2) = f(a) + f(b^2) + ab^2.
$$
Given that $f(40) = 2020$ find $f(1)$.
$$
f(a + b^2) = f(a) + f(b^2) + ab^2.
$$
Given that $f(40) = 2020$ find $f(1)$.