Problem
ALG-B3-M04-P004 Surjectivity Without Extra Substitutions
#4
★★☆☆☆ Level 2 of 5
Find all surjective \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=x+f(y)+1\) for all \(x,y\).
Again put \(t=f(y)\).
Since \(f\) is surjective, \(t=f(y)\) runs through all real numbers. Hence \(f(x+t)=x+t+1\) for all \(x,t\). Every number \(u\) can be written as \(u=x+t\), so \(f(u)=u+1\). The check is immediate: \(f(x+f(y))=x+f(y)+1\). The answer is \(f(x)=x+1\).
A useful short task showing that the pair \((x,t)\) creates an arbitrary sum.