Problem
ALG-B3-M11-P018 Modular Function
#18
★★★★★ Level 5 of 5
Let \(p\) be an odd prime, \(f:\mathbb Z/p\mathbb Z\to\mathbb Z/p\mathbb Z\), and \(f(x+f(y))=f(x)+y\). Find \(f\).
Prove injectivity, then additivity.
With \(x=0\), \(f(f(y))=f(0)+y\), so \(f\) is injective, hence bijective on a finite set. With \(y=0\), \(f(x+f(0))=f(x)\), so \(f(0)=0\). Thus \(f(f(y))=y\). Putting \(y=f(t)\), we get \(f(x+t)=f(x)+f(t)\), hence \(f(x)=ax\). From \(a^2=1\) modulo \(p\), \(a=\pm1\). The answer is \(f(x)=x\), \(f(x)=-x\).
Mix of finiteness and injectivity.