Problem
ALG-B3-M04-P010 The Extra One
#10
★★★☆☆ Level 3 of 5
Prove that there is no surjective function \(f:\mathbb R\to\mathbb R\) such that \(f(x+f(y))=f(x)+f(y)+1\) for all \(x,y\).
After replacing \(f(y)=t\), make sure to substitute the resulting form back.
By surjectivity, \(t=f(y)\) is arbitrary, so \(f(x+t)=f(x)+t+1\). Taking \(x=0\), we get \(f(t)=f(0)+t+1\). Then the left-hand side of \(f(x+t)=f(x)+t+1\) equals \(f(0)+x+t+1\), while the right-hand side equals \(f(0)+x+1+t+1\). This gives \(0=1\), a contradiction. No such function exists.
A good checking task: an incomplete verification gives a false affine candidate.