Problem
ALG-B3-M08-P017 Negative Square
#17
★★★★★ Level 5 of 5
Prove that there is no additive \(f:\mathbb Q\to\mathbb Q\) such that \(f(f(x))=-x\) for all \(x\).
On \(\mathbb Q\), we would have \(f(x)=cx\).
Additivity on \(\mathbb Q\) gives \(f(x)=cx\) with rational \(c\). Then \(f(f(x))=c^2x\). The condition requires \(c^2=-1\), impossible for a rational number. Hence no such function exists.
Impossibility from the iteration coefficient.