Problem
ALG-B3-M01-P020 Why verification is mandatory
#20
★★★★★ Level 5 of 5
For \(f:\mathbb R\to\mathbb R\), consider \(f(x+y)=f(x)+f(y)+xy\). Prove that if \(f(x)=\frac{x^2}{2}+A(x)\), where \(A\) is additive, then \(f\) is a solution. Explain why over \(\mathbb R\) this is not restricted to polynomials.
Hint. Substitute the formula and use \(A(x+y)=A(x)+A(y)\).
We have \(f(x+y)=\frac{(x+y)^2}{2}+A(x+y)=\frac{x^2}{2}+\frac{y^2}{2}+xy+A(x)+A(y)=f(x)+f(y)+xy\). If \(A\) is an arbitrary additive function on \(\mathbb R\), it need not be linear without regularity assumptions. Thus the solution set is broader than just quadratic polynomials.
Goal: separate guessing the answer from a complete proof and domain check.