Problem
ALG-B3-M04-P015 The Golden Ratio in a Functional Equation
Find all continuous \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=f(x)+f(y)+y\) for all \(x,y\).
After the shift \(g(x)=f(x)-f(0)\), an additive function should appear.
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 \(f(x+t)=f(x)+f(t)-c\). Therefore \(g(x)=f(x)-c\) is additive and continuous, so \(g(x)=ax\).
Thus \(f(x)=ax+c\). Substituting into the original equation gives \(a(x+ay+c)+c=ax+(a+1)y+2c\). Hence \(a^2=a+1\) and \(c(a-1)=0\). Since the roots of the first equation are not \(1\), we get \(c=0\). Thus \(a=\frac{1+\sqrt{5}}{2}\) or \(a=\frac{1-\sqrt{5}}{2}\). Both functions \(f(x)=ax\) work.
This is an olympiad-style chain: injectivity, shift, additivity, coefficient comparison.