Problem
ALG-B3-M04-P007 A Preimage Inside the Argument
#7
★★★☆☆ Level 3 of 5
Find all \(f:\mathbb Q\to\mathbb Q\) satisfying \(f(f(x)+y)=x+f(y)\) for all \(x,y\in\mathbb Q\).
Prove \(f(0)=0\), then substitute \(y=f(t)\).
With \(y=0\), \(f(f(x))=x+f(0)\), so \(f\) is injective and surjective. With \(x=0\), \(f(f(0)+y)=f(y)\). By injectivity, \(f(0)+y=y\), hence \(f(0)=0\). Thus \(f(f(x))=x\).
Put \(y=f(t)\): \(f(f(x)+f(t))=x+t=f(f(x+t))\). By injectivity, \(f(x)+f(t)=f(x+t)\). Hence \(f\) is additive on \(\mathbb Q\), so \(f(x)=cx\). The condition \(f(f(x))=x\) gives \(c^2=1\). The answer is \(f(x)=x\) or \(f(x)=-x\).
Similar to the previous one, but injectivity and surjectivity come from a different substitution.