Problem
ALG-B3-M04-P018 A Hidden Inverse Function
Find all increasing functions \(f:\mathbb R\to\mathbb R\) such that \(f(f(x)+f(y))=x+y\) for all \(x,y\).
First prove bijectivity, then compare \(f\) with its inverse.
If \(f(a)=f(b)\), then using the equations with the same \(y\) gives \(a+y=b+y\), so \(a=b\). Thus \(f\) is injective. Also, every \(T\) equals \(f(f(T)+f(0))\), so \(f\) is surjective. Let \(g=f^{-1}\).
Putting \(y=0\), we get \(f(f(x)+f(0))=x\), hence \(g(x)=f(x)+f(0)\). Let \(c=f(0)\). Then \(g(x)=f(x)+c\). From the original equation with \(x=y=g(0)\), we obtain \(f(0)=2g(0)\). But from \(g(0)=f(0)+c\), we get \(g(0)=2c\). Hence \(c=4c\), so \(c=0\). Therefore \(g=f\) and \(f(f(x))=x\).
Now for arbitrary \(u=f(x)\), \(v=f(y)\), we have \(f(u+v)=x+y=f(u)+f(v)\), because \(g=f\). Thus \(f\) is additive. An increasing additive function has the form \(f(x)=ax\), where \(a>0\). The condition \(f(f(x))=x\) gives \(a^2=1\), hence \(a=1\). The answer is \(f(x)=x\).
A harder problem with two ideas: build the inverse function and then derive additivity.