Problem
ALG-B3-M02-P018 Injectivity from a shift
#18
★★★★★ Level 5 of 5
Find all \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=x+2y\) and \(f\) is surjective.
Hint. Compare the relations after substituting \(x=0\) and use surjectivity.
With \(x=0\), \(f(f(y))=2y\). Since \(f\) is surjective, for every \(t\) there is \(y\) with \(t=f(y)\), and then \(f(t)=2y\). Hence \(f(x+t)=x+f(t)\) for all \(x,t\). Thus \(f(u)-u\) is constant: \(f(u)=u+c\). Then \(f(f(y))=y+2c\), but it must be \(2y\), impossible for all \(y\). Therefore there are no solutions.
Goal: choose the right first substitution and then verify the found function.