Mathematical Reflections 2017, Issue 4 - Problem S416

Problem:
Let $f:\mathbb{N} \to \{\pm 1\}$ be a function such that $f(mn)=f(m)f(n)$ for all $m,n \in \mathbb{N}$. Prove that
there are infinitely many $n$ such that $f(n)=f(n+1)$.

Proposed by Oleksi Klurman, University College London, UK

Solution:
Let $n \in \mathbb{N}$. As $f$ is completely multiplicative, then $f(n^2)=(f(n))^2=1$. If there are infinitely many $n$ such that $f(n^2-1)=1$, we are done. So, assume that there are only finitely many $n$ such that $f(n^2-1)=1$. Then, there are infinitely many $n$ such that $f(n^2-1)=-1$. Since $f(n^2-1)=f(n-1)f(n+1)$, then there are infinitely many pairs $\{f(n-1),f(n+1)\}=\{-1,1\}$. Since $f(n)=\pm 1$, it follows that there are infinitely many $n$ such that $f(n)=f(n+1)$.