Baltic Way shortlist

It is not difficult to find $f(n)$ for small $n$: $$f(1\cdot 1)=f(1)f(1)\text{ therefore }f(1)=1;$$ $$f(3)=f(1)+f(1)+f(1)=3;$$ $$f(5)=f(1+1+3)=5;$$ $$f(10)=f(1+3+6)=4+3f(2)\text{ and }f(10)=f(2\cdot 5)=f(2)f(5)=5f(2)\text{ therefore }f(2)=2.$$ Now we use induction. Suppose that $f(n)=n$ for all $n<N$. Let us show that $f(N)=N$. Since $f$ is multiplicative we may assume that $N=p^r$ for some prime $p$. Consider several similar cases.

1) $N=3^r$. Then $$f(3T_{3^{r-1}})=3f(T_{3^{r-1}})=3f\left(\frac{3^{r-1}(3^{r-1}+1)}{2}\right)=3f(3^{r-1})f\left(\frac{3^{r-1}+1}{2}\right)$$ And from the other hand $$f(3T_{3^{r-1}})=f\left(\frac{3^r(3^{r-1}+1)}{2}\right)=f(3^r)f\left(\frac{3^{r-1}+1}{2}\right)$$ So we conclude that $f(3^r)=3^r$ since $f(3^{r-1})=3^{r-1}$ by induction hypothesis.

and $$f(T_{s-1}+T_{s-1}+T_s)=f\left(\frac{s(3s-1)}{2}\right)=f\left(\frac{s}{2}\right)f(3s-1)=\frac{s}{2}f(p^r).$$ Hence $f(p^r)=p^r$.

3) $N=p^r$, where $p$ is an odd prime and $p^r=3s+1$. Similarly we have $$\begin{array}{rl}f(T_{s-1}+T_s+T_s)& =\frac{s(s-1)}{2}+\frac{s(s+1)}{2}+\frac{s(s+1)}{2}=\frac{s(3s+1)}{2}=\frac{s{p}^r}{2}\\ & =f\left(\frac{s(3s+1)}{2}\right)=f\left(\frac{s}{2}\right)f(3s+1)=\frac{s}{2}f(p^r).\end{array}$$ Hence $f(p^r)=p^r$.