Problem
ALG-B3-M05-P019 The Unit Shift
#19
★★★★★ Level 5 of 5
Find all \(f:\mathbb Q\to\mathbb Q\) such that \(f(x+y)=f(x)+f(y)-1\) and \(f(xy)=f(x)f(y)-f(x)-f(y)+2\).
After the substitution \(g=f-1\), both additivity and multiplicativity should appear.
Let \(g(x)=f(x)-1\). Then \(g(x+y)=g(x)+g(y)\). The second condition gives \(g(xy)+1=(g(x)+1)(g(y)+1)-(g(x)+1)-(g(y)+1)+2=g(x)g(y)+1\), hence \(g(xy)=g(x)g(y)\). On \(\mathbb Q\), additivity gives \(g(x)=cx\). Multiplicativity gives \(c=c^2\), so \(c=0\) or \(c=1\). The answer is \(f(x)=1\) and \(f(x)=x+1\).
Repeats an important harder template with full solution.