Functional Equation Problem, find $f(1998)$ given certain statements...

Question

Find $f(1998)$ and prove you have found the right answer given that:

• There is a $f(n)$ for every $n$;
• $f(n)$ is a whole number
• $f(2) = 2$
• $f(mn) = f(m)f(n)$
• $f(m)> f(n)$ if $m>n$

What I've done so far:

• Proved $f(1) = 1$

  • let $m=2, n=1$
  • $f(2)=f(2) * f(1)$
  • $2 = 2 * f(1) $
  • $f(1) = 1$

• Guessing that $f(x) = x$ ..?

Thanks!

Answer 1

Ok let's see what we can get:

• we know already $f(1)=1$, $f(2)=2$
• $f(4) = f(2\cdot 2) = f(2)\cdot f(2) = 2\cdot 2 = 4$
• $2<3<4 \implies f(2)<f(3)<f(4) \implies 2<f(3)<4 \implies f(3) = 3$
• $f(6) = f(2\cdot 3) = f(2)\cdot f(3) = 2\cdot 3 = 6$
• $4<5<6 \implies f(4)<f(5)<f(6) \implies 4<f(5)<6 \implies f(5) = 5$
• $f(8) = f(2\cdot 4) = f(2)\cdot f(4) = 2\cdot 4 = 8$
• $6<7<8 \implies f(6)<f(7)<f(8) \implies 6<f(7)<8 \implies f(7) = 7$
• ...

It appears that we could show by induction that $f(2k+1) = 2k+1$, $f(2k+2) = 2k+2$ for all $k\in\mathbb{N}$.

$$\implies f(1998) = 1998$$

Answer 2

Note that $f(n^m)=(f(n))^m$. We claim that $f(n)=n$.For the sake of contradiction, let's assume $\exists a \in \Bbb N$,such that $f(a) \neq a$. Let's assume that $f(a) \geq a+1$ (other case is similiar). Choose a rational number $\frac mn$ such that $\log_2a< \frac mn<\log_2(a+1)$.Then, $$ a^n<2^m<(a+1)^n \leq f(a^n)\\ \implies f(2^m)<f(a^n) \quad and\ \ 2^m>a^n $$ This is a contradiction as $f(n)$ is strictly increasing. We can arrive at a similiar contradiction if $f(a)<a$. Therefore we conclude that $f(n)=n \quad \forall n\in \Bbb N.$