Problem
ALG-B3-M04-P009 An Impossible Squared Coefficient
Prove that there is no function \(f:\mathbb Q\to\mathbb Q\) such that \(f(x+f(y))=f(x)+3y\) for all \(x,y\in\mathbb Q\).
Reduce the equation to additivity and to the condition \(c^2=3\).
If \(f(a)=f(b)\), comparing the original equality for \(y=a\) and \(y=b\) gives \(3a=3b\), so \(a=b\). Thus \(f\) is injective. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). With \(x=0\), \(f(f(y))=3y\), so \(f\) is surjective.
Let \(t=f(y)\). Then \(f(t)=3y\), and the original equation becomes \(f(x+t)=f(x)+f(t)\). Hence \(f\) is additive on \(\mathbb Q\), so \(f(x)=cx\) for some rational \(c\). The condition \(f(f(y))=3y\) gives \(c^2=3\), impossible for rational \(c\). Contradiction.
Important because the answer is not found; impossibility appears only after the full method.