Problem
ALG-B3-M04-P016 A Surjective Composition
#16
★★★★★ Level 5 of 5
Find all surjective functions \(f:\mathbb R\to\mathbb R\) such that \(f(f(x)+y)=f(x)+f(y)\) for all \(x,y\).
Let \(t=f(x)\) be arbitrary.
Since \(f\) is surjective, \(t=f(x)\) runs through all real numbers. Then \(f(t+y)=t+f(y)\) for all \(t,y\). Taking \(y=0\), we obtain \(f(t)=t+f(0)\). Hence \(f(x)=x+c\). Checking: \(f(f(x)+y)=f(x)+y+c=x+c+y+c=f(x)+f(y)\). All functions \(f(x)=x+c\) work.
Surjectivity is applied to a different argument, not to \(f(y)\).