Summation of a strictly increasing and completely multiplicative function satisfying $f(2)=2$

Question

Let $f:\mathbb N\to\mathbb N$ be a function satisfying the following conditions:

1. $f(mn)=f(m)\times f(n)$;
2. $f(m)<f(n)$ if $m<n$;
3. $f(2)=2$.

What is the value of $\sum_{i=1}^{20}f(i)$?

I intuitively guessed that the identity function satisfied the above conditions and solved the question using that, but am unable to solve it generally. I need help proving that there was only one unique function satisfying the criteria. What is the solution to the problem?

Answer 1

Hint: $f(2)=f(1\times2)=f(1)f(2) \implies f(1)=1$. Now $2=f(2) \lt f(3) \lt f(4)=4$ giving $f(3)=3$. This gives that $f(6)=6$ and using $f(4) \lt f(5) \lt f(6)$ one can conclue that $f(5)=5$. Going on like this you can show that $f(m)=m$ for every $m$. It is now easy to calculate what you need.