Problem
ALG-B3-M04-P006 Two Involutions on the Rationals
#6
★★★☆☆ Level 3 of 5
Find all functions \(f:\mathbb Q\to\mathbb Q\) such that \(f(x+f(y))=f(x)+y\) for all \(x,y\in\mathbb Q\).
First obtain \(f(f(y))=f(0)+y\), then prove injectivity.
With \(x=0\), \(f(f(y))=f(0)+y\). If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a=b\). Thus \(f\) is injective. With \(y=0\), \(f(x+f(0))=f(x)\), so \(f(0)=0\). Hence \(f(f(y))=y\).
Put \(y=f(t)\): \(f(x+f(f(t)))=f(x)+f(t)\), so \(f(x+t)=f(x)+f(t)\). On \(\mathbb Q\), an additive function has the form \(f(x)=cx\). From \(f(f(y))=y\), we get \(c^2=1\). The answers are \(f(x)=x\) and \(f(x)=-x\). Direct checking works.
A standard module chain: injectivity, zero, additivity, verification.