Problem
ALG-B3-M02-P016 Product inside the function
#16
★★★★★ Level 5 of 5
Find all \(f:\mathbb Q\to\mathbb Q\) such that \(f(xy)=x f(y)+y f(x)\) for all \(x,y\in\mathbb Q\) and \(f(2)=0\).
Hint. Substitute \(y=1\), then \(x=2\).
With \(y=1\), \(f(x)=x f(1)+f(x)\), so \(f(1)=0\). Put \(f(x)=xD(x)\) for \(x\ne0\). Then the equation becomes \(xyD(xy)=xyD(y)+xyD(x)\), hence \(D(xy)=D(x)+D(y)\), and \(D(2)=0\). Conversely, every multiplicatively additive \(D:\mathbb Q^{*}\to\mathbb Q\) with \(D(2)=0\), together with \(f(0)=0\), gives a solution by \(f(x)=xD(x)\). Thus the answer is a family, not a single formula.
Goal: choose the right first substitution and then verify the found function.