What is the least possible value of $(999)?$ Let $$ be a one-to-one function from the set of natural numbers to itself such that $() = ()()$ for all natural numbers $$ and $.$ What is the least possible value of $(999)?$
What I think is that I have a hunch for the answer and follow the intuitions, any help?
 A: Very nice problem, I have solved it before I don't know where, and this is my solution:
Put $m=1\Rightarrow f(n)=f(1)f(n)\Rightarrow  f(1)=1$ as $f(n)\neq 0$. As $f(x)$ is one-one function $\Rightarrow f(n)> 1,   \forall n> 1$.
Now $f(999)=f(3)^{3}f(37)\geq 2^{3}\cdot 3=24.$
We claim that equality does occur at $24$ by setting $f(3)=2, f(2)=37, f(37)=3, f(p)=p$ $\forall$ primes $p\neq 2,3,37$ and $f(x)$ to be obtained from $f(mn) =f(m)f(n)$, whenever $x$ is not prime. $f(999)$ will obtain its least value for this function once it is justified that the function is one-one. Now assume $f(N)=f(M)$ and suppose the prime factorization of $N$ and $M$ be $2^{x_{1}} 3^{x_{2}} 5^{x_{3}}...37^{x_{12}}... $ and $2^{y_{1}} 3^{y_{2}} 5^{y_{3}}...37^{y_{12}}...$ $$\Rightarrow f(N)=f(2^{x_{1}} 3^{x_{2}} 5^{x_{3}}...37^{x_{12}}...)=f(2)^{x_{1}} f(3)^{x_{2}}...f(37)^{x_{12}}...=37^{x_{1}} 2^{x_{2}}...3^{x_{12}}$$
and similarly  $$ \Rightarrow f(M)=f(2^{y_{1}} 3^{y_{2}} 5^{y_{3}}...37^{y_{12}}...)=f(2)^{y_{1}} f(3)^{y_{2}}...f(37)^{y_{12}}...=37^{y_{1}} 2^{y_{2}}...3^{y_{12}}$$ $$f(N)=f(M)\Rightarrow x_{1}=y_{1}, x_{2}=y_{2},...,x_{12}=y_{12}\Rightarrow N=M$$
AND WE'RE DONE
