Problem
ALG-B3-M04-P019 A Condition Killing Both Answers
#19
★★★★★ Level 5 of 5
Prove that there is no continuous \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=f(x)+y\) for all \(x,y\) and \(f(2)=3\).
First solve the equation without the condition \(f(2)=3\).
With \(x=0\), \(f(f(y))=f(0)+y\). This implies injectivity in the usual way. With \(y=0\), \(f(x+f(0))=f(x)\), hence \(f(0)=0\). Thus \(f(f(y))=y\).
Putting \(y=f(t)\), we get \(f(x+t)=f(x)+f(t)\), so \(f\) is additive. A continuous additive function has the form \(f(x)=cx\). From \(f(f(y))=y\), we get \(c^2=1\), hence \(f(x)=x\) or \(f(x)=-x\). In the first case \(f(2)=2\), in the second \(f(2)=-2\). Neither gives \(3\). Contradiction.
Useful as a complete-solution problem, not a local substitution at the given point.