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Algebra Difficulty 4.8 AIME Prove it Ireland

Let N\mathbb{N} denote the set of all non-negative integers (including zero). A function f:N→Nf: \mathbb{N} \to \mathbb{N} satisfies for all a,b∈Na, b \in \mathbb{N}
f(a+b2)=f(a)+f(b2)+ab2. f(a + b^2) = f(a) + f(b^2) + ab^2.
Given that f(40)=2020f(40) = 2020 find f(1)f(1).

Solution

Putting b=1b = 1 in the recurrence relation, we have: f(a+1)=f(a)+f(1)+af(a+1) = f(a) + f(1) + a. It is then an easy inductive proof that for a≥1a \ge 1
f(a)=a(f(1)+a−12). f(a) = a \left( f(1) + \frac{a-1}{2} \right).
On substitution, this satisfies the functional equation for any a,b∈Na, b \in \mathbb{N} and so all solutions are of this form. Given f(40)=2020f(40) = 2020 we must have:
2020=40(f(1)+392). 2020 = 40 \left( f(1) + \frac{39}{2} \right).
This implies f(1)=31f(1) = 31.

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