Problem
ALG-B3-M04-P005 A Sum Inside the Function
#5
★★☆☆☆ Level 2 of 5
Let \(f:\mathbb R\to\mathbb R\) be injective and let \(f(f(x)+f(y))=f(x+y)\) for all \(x,y\). Find \(f\).
Cancel the outside \(f\), then put \(y=0\).
By injectivity, \(f(x)+f(y)=x+y\). Taking \(y=0\), we get \(f(x)+f(0)=x\), so \(f(x)=x-c\), where \(c=f(0)\). Substitute this into \(f(x)+f(y)=x+y\): \((x-c)+(y-c)=x+y\), hence \(c=0\). Therefore \(f(x)=x\). The check is immediate.
The student should prove the identity function through injectivity, not guess it.