Strictly increasing functions $f$ with $f(mn) = f(m) + f(n) + f(m)f(n)$ , $f(2) = 7$

Question

Find all (strictly) increasing functions $f$ from the nonnegative integers to the integers satisfying $f(2) = 7$ and $$f(mn) = f(m) + f(n) + f(m)f(n)$$ for all nonnegative integers $m$ and $n$.

My Progress: By $(m,n)=(2,0)$ , we get $f(0)=-1$ .

By $(m,n)=(2,1)$ , we get $f(1)=0 $ .

And then by induction, we can show $f(2^x)=8^x-1$ . So $f(1)=0$, $f(2)=7$,$ f(4)=63$, $f(8)= 511$ ,$ f(16)=4095$ , $f(32)=32767$, $f(64)=262143$ , $f(128)=2097151$ , $f(256)=16777215$, $f(512)=134217727$, $f(1024)=1073741823$

Then I tried to find $f(3)$ . Could get anything nice though

Now, I took $f(3)= 10$, then $f(243)<f(128)$ . Hence $f(3)>10$. then when I took $f(3)=40$ , then $f(81)>f(128)$ . Hence $f(3)<40 $. then I took $f(3)=25$, then $f(243)>f(256)$. hence $\boxed {f(3)<25}$. then I took $f(3)=15$ , then $f(27)=f(16)$. hence $f(3)> 15$ . then I took $f(3)=20$, then $f(81)<f(64)$. Hence $\boxed{f(3)>20}$.

And then I stopped, since , I highly feel, I am in a wrong path ... So can someone give me hints?

Thanks in advance .

Answer 1

Since $g(x)=f(x)+1$ is completely multiplicative, it's enough to find the values $g(p)$ for any prime $p$. We show that $g(x)=x^3$. By the multiplicativity, it suffices to prove that $g(p)=p^3$ for all primes $p$. Assume, for the sake of contradiction, that $g(p)\geq p^3+1$ for some prime $p$ (the other case is analogous). Choose a positive rational number $\alpha = \frac{m}{n}$ such that $p^3 < 2^{\alpha} < p^3+1$, this is possible since numbers of these form are obviously dense in $(0, \infty)$. Then we get $p^{3n} < 2^m$ and thus, since $g$ is increasing, $$g(p)^{3n} = g(p^{3n}) < g(2^m)=2^{3m} \Longrightarrow g(p) < 2^{\alpha}.$$ On the other hand, $g(p) \geq p^3+1 > 2^{\alpha}$. by assumption. This is a contradiction, thus we get $g(x)=x^3$ and $f(x)=x^3-1$ for all nonnegative integers $x$.

Remark: It's easy to see that if $f(2)$ is not specified, then there are infinitely many solutions to the functional equation $$f(mn)=f(m)+f(n)+f(m)f(n)$$ We can just assign to $g(p)$ values as we wish for primes $p$. So it will be a nice problem to investigate when we fix $f(2)=q$ for some other primes. So there are some natural question,

1. Do we always have a unique function if we fix a prime value of $f(2)$?
2. Does there exist primes $q$ such that if we fix $f(2)=q$ then we can get more than one functions $f$? (Is there something special about $7$?).

Answer 2

With @AlexeyBurdin's hint $g(n):=f(n)+1$ is a (strictly) increasing totally multiplicative function with $g(n)=n^3$ whenever $n=0$ or $n$ is a power of $2$ (including $n=2^0=1$). For any other $n\ge3$, $n$ is arbitrarily well-approximated by $2^q$ with $q\in\Bbb Q$, so $g(n)$ is arbitrarily well-approximated for such $q$ by $2^{3q}$; since $g$ is increasing, $f(n)=n^3-1$ is the only function. In particular, say rational sequences $a_k/b_k,\,c_k/d_k$ respectively approach $\log _2n$ from below and from above, with $a_k,\,\cdots,\,d_k$ positive integers; then$$2^{a_k}\le n^{b_k}\implies 8^{a_k}\le g(n)^{b_k}\implies g(n)\ge8^{a_k/b_k},$$and similarly $g(n)\le8^{c_k/d_k}$.