Problem
ALG-B3-M06-P014 A Preimage on Integers
#14
★★★★★ Level 5 of 5
Find all \(f:\mathbb Z\to\mathbb Z\) such that \(f(n+f(m))=f(n)+m\) for all integers \(m,n\).
First prove \(f(0)=0\) and \(f(f(m))=m\).
With \(n=0\), \(f(f(m))=f(0)+m\). If \(f(a)=f(b)\), then \(f(f(a))=f(f(b))\), hence \(a=b\); so \(f\) is injective. With \(m=0\), \(f(n+f(0))=f(n)\), hence \(f(0)=0\). Thus \(f(f(m))=m\). Put \(m=f(t)\): \(f(n+t)=f(n)+f(t)\). Therefore \(f\) is additive on \(\mathbb Z\), so \(f(n)=cn\). From \(f(f(n))=n\), we get \(c^2=1\). The answer is \(f(n)=n\) and \(f(n)=-n\).
Connects this module with the previous one on bijectivity.