Problem
ALG-B3-M04-P012 A Golden Equation Without a Rational Answer
Prove that there is no \(f:\mathbb Q\to\mathbb Q\) such that \(f(x+f(y))=f(x)+f(y)+y\) for all \(x,y\in\mathbb Q\).
Again try to obtain \(f(x+t)=f(x)+f(t)-f(0)\).
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 \(x=0\), \(f(f(y))=c+f(y)+y\). If \(t=f(y)\), then \(f(t)=c+t+y\), so \(t+y=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Thus \(g(x)=f(x)-c\) is additive, and on \(\mathbb Q\), \(g(x)=ax\).
Therefore \(f(x)=ax+c\). Substitute into the original equation: \(a(x+ay+c)+c=ax+(a+1)y+2c\). Comparing coefficients gives \(a^2=a+1\) and \(c(a-1)=0\). But \(a^2-a-1=0\) has no rational roots. Contradiction.
The same architecture, but the answer is impossible because of the domain.