67th NMO Selection Tests for BMO and IMO

To this end, write $\ell =f(1)$, and notice that, since $1\le f(n)/n\le \ell$, there exists a minimal positive integer $k\le \ell$ such that $\lfloor f(n)/n\rfloor =k$ for infinitely many positive integers $n$. Let $A$ denote the set of all these positive integers, let $B=\{n:n\in A\text{ and }2n\notin A\}$, and let $A^{\prime }=A\setminus B$.

We now show that $B$ is finite, so $A^{\prime }$ is infinite. Indeed, $f(2n)\le 2f(n)$ and $\lfloor f(2n)/(2n)\rfloor \le \lfloor 2f(n)/(2n)\rfloor =k$ imply $\lfloor f(2n)/(2n)\rfloor <k$ for all $n$ in $B$, so $B$ is finite by minimality of $k$.

Next, for $n$ in $A^{\prime }$, write $$\frac{2f(n)}{f(2n)}=\frac{\frac{f(n)}{n}}{\frac{f(2n)}{2n}}<1+\frac{1}{k},$$ to deduce that the positive integer $2f(n)/f(2n)$ is less than $2$, so $f(2n)=2f(n)$.

Further, fix a positive integer $a$ and notice, as before by minimality of $k$, that $f(a+n)\ge k(a+n)$ for all but finitely many $n$ in $A^{\prime }$. Hence $$\frac{f(a)+f(n)}{f(a+n)}<\frac{f(a)+(k+1)n}{k(a+n)},$$ for all but finitely many $n$ in $A^{\prime }$, so $(f(a)+f(n))/f(a+n)$ is a positive integer less than $2$ for all but finitely many $n$ in $A^{\prime }$, i.e., $f(a+n)=f(a)+f(n)$ for all but finitely many $n$ in $A^{\prime }$.

Finally, fix two positive integers $a$ and $b$. By the preceding, $f(a+n)=f(a)+f(n)$, $f(b+n)=f(b)+f(n)$, and $(f(a+n)+f(b+n))/f(a+b+2n)$ is a positive integer for all but finitely many $n$ in $A^{\prime }$. Consequently, $$\frac{f(a)+f(b)+2f(n)}{f(a+b)+f(2n)}=\frac{f(a)+f(b)+f(2n)}{f(a+b)+f(2n)}$$ is a positive integer for all but finitely many $n$ in $A^{\prime }$, so the latter must equal $1$ for all but finitely many $n$ in $A^{\prime }$, i.e., $f(a+b)=f(a)+f(b)$. This ends the proof.