Re: Functional Equation (Canada 1969)
Posted: Mon Sep 22, 2014 8:13 am
No. The three conditions
$f:\mathbb N \mapsto \mathbb N, f(mn)=f(m)f(n)$
$f$ strictly increasing.
$f(2)=2$ imply the unique solution $f(n)=n$.
$f:\mathbb N \mapsto \mathbb N, f(mn)=f(m)f(n)$
$f$ strictly increasing.
$f(2)=2$ imply the unique solution $f(n)=n$.