Problem
ALG-B3-M10-P016 Surjective Form with a Coefficient
#16
★★★★★ Level 5 of 5
Find all increasing \(f:\mathbb R\to\mathbb R\) such that \(f(f(x)+y)=x+f(y)\).
Prove \(f(0)=0\), then additivity.
With \(y=0\), \(f(f(x))=x+f(0)\), which implies injectivity and surjectivity. With \(x=0\), \(f(f(0)+y)=f(y)\), so \(f(0)=0\). Thus \(f(f(x))=x\). Putting \(y=f(t)\), we get \(f(f(x)+f(t))=x+t=f(f(x+t))\), hence \(f(x)+f(t)=f(x+t)\). The function is additive and increasing, so \(f(x)=cx\). From \(f(f(x))=x\), \(c^2=1\), and increasingness gives \(c=1\). The answer is \(f(x)=x\).
Strong mixed problem using several modules.