Problem
ALG-B3-M04-P020 The Golden Shift
Find all continuous functions \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=f(x)+f(y)+y+1\) for all \(x,y\).
Use the shift \(g(x)=f(x)-f(0)\), but keep careful track of the free constant \(1\).
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 \(x=0\), \(f(f(y))=c+f(y)+y+1\). Put \(t=f(y)\). Then \(f(t)=c+t+y+1\), so \(t+y+1=f(t)-c\). Hence \(f(x+t)=f(x)+f(t)-c\). Therefore \(g(x)=f(x)-c\) is additive and continuous, so \(g(x)=ax\).
From \(f(c)=f(0+f(0))=c+c+0+1=2c+1\), we get \(g(c)=c+1\). Also, \(f(f(y))=c+f(y)+y+1\) becomes \(g(g(y)+c)+c=g(y)+y+2c+1\). Since \(g\) is additive and \(g(c)=c+1\), this gives \(g(g(y))=g(y)+y\). For \(g(x)=ax\), we get \(a^2=a+1\).
The remaining condition is \(g(c)=c+1\), that is \(ac=c+1\), or \(c(a-1)=1\). Thus \(a\) is one of the roots of \(a^2-a-1=0\), and \(c=\frac{1}{a-1}\). Hence the two functions are \(f(x)=ax+\frac{1}{a-1}\), where \(a=\frac{1+\sqrt{5}}{2}\) or \(a=\frac{1-\sqrt{5}}{2}\). Direct substitution confirms both.
Final problem of the stage: the same architecture, but the constant does not disappear and must be tracked carefully.