Problem
ALG-B3-M04-P011 A Shift After an Involution
Find all functions \(f:\mathbb Q\to\mathbb Q\) such that \(f(x+f(y))=f(x)+y+1\) for all \(x,y\in\mathbb Q\).
Prove that \(g(x)=f(x)-f(0)\) is additive.
If \(f(a)=f(b)\), comparing the equations for \(y=a\) and \(y=b\) gives \(a=b\), so \(f\) is injective. Let \(c=f(0)\). With \(x=0\), \(f(f(y))=c+y+1\). Put \(t=f(y)\). Then \(f(t)=c+y+1\), so \(y+1=f(t)-c\). Hence the original equation gives \(f(x+t)=f(x)+f(t)-c\) for all \(x,t\) in the image. The function is surjective because \(f(f(T-c-1))=T\), so this holds for all \(t\).
Let \(g(x)=f(x)-c\). Then \(g(x+t)=g(x)+g(t)\), so on \(\mathbb Q\) we have \(g(x)=ax\). Also, from \(f(f(y))=c+y+1\) we get \(g(g(y)+c)=y+1\). Since \(g\) is additive, \(g(g(y))+g(c)=y+1\). At \(y=0\), \(g(c)=1\), hence \(g(g(y))=y\). Therefore \(a^2=1\) and \(ac=1\). If \(a=1\), then \(c=1\); if \(a=-1\), then \(c=-1\). The answer is \(f(x)=x+1\) and \(f(x)=-x-1\). Direct checking works.
The level rises because the function has to be shifted by \(f(0)\) correctly.