Problem
ALG-B3-M01-P015 Injectivity from the equation
#15
★★★★★ Level 5 of 5
Find all \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=x+y\) for all \(x,y\in\mathbb R\).
Hint. First prove that \(f\) is injective.
If \(f(u)=f(v)\), then \(f(x+f(u))=x+u\) and \(f(x+f(v))=x+v\) imply \(u=v\). Thus \(f\) is injective. With \(x=0\), \(f(f(y))=y\). Substitute \(y=f(t)\): \(f(x+t)=x+f(t)\). For fixed \(t\), this gives \(f(s)=s+f(t)-t\), so \(f(s)=s+c\). Then \(f(f(y))=y+2c\), hence \(c=0\). Answer: \(f(x)=x\).
Goal: separate guessing the answer from a complete proof and domain check.