increasing sequence specific properties I am doing some olympiad exercises and have difficulties with the following one:
Consider a sequence $a_1,a_2,...$ which is strictly monotonically increasing and $a_1,a_2,...\in\mathbb N$
Now I know that $a_2=2$ and $a_{xy}=a_xa_y$ if $\gcd(x,y)=1$
I would like to show that $a_n=n$
The first thing I did was to write down the first values $a_1,2,a_3,a_4,a_5,2a_3,...$
Now I think it is necessary to separate cases where n is odd and even but I have no idea how to do it, may you could help me.
EDIT: As discussed in the comments, it seems the sequence $a_n=n$ is not unique, but I would like to prove $a_n=n$ using my assumptions, no matter if it is unique or not. If you see a different explicit formula I would also be interested in.
 A: Let $f(x)=a_x$. The strict monotonicity hypothesis can be rewritten as 
$f(j)-f(i) \geq j-i$ for any $i<j$.
Denote by $\Omega$ the set of all integers $x \geq 1$ such that
$f(x)=x$. Since $f$ is strictly monotonic,  $\Omega$ is “convex” :
if $a<b$ are in $\Omega$ then all the integers between $a$ and $b$ 
are in $\Omega$ also.
Also, $\Omega$ is weakly multiplicative : if $a,b\in \Omega$ are coprime,
then $ab\in \Omega$.
We have $f(15)=f(3)f(5)$ and $f(18)=2f(9)$, so $2f(9) \geq f(3)f(5)+3$ by
strict monotonicity. On the other hand, $f(10)=2f(5)$ and hence
$2f(5) \geq f(9)+1$ by strict monotonicity. Combining those two inequalities,
$$
4f(5) \geq 2f(9)+2 \geq f(3)f(5)+5 \gt f(3)f(5)
$$
This implies $4 \gt f(3)$, so $f(3)=3$ and hence $3\in \Omega$.
So $6=2\times 3 \in \Omega$ since $\Omega$ is weakly multiplicative.
Then $2\in \Omega,6\in \Omega$ yields $[2,6] \subseteq \Omega$ since $\Omega$ is convex.
So $15=3\times 5 \in \Omega$ since $\Omega$ is weakly multiplicative.
Then $2\in \Omega,15\in \Omega$ yields $[2,15] \subseteq \Omega$ since $\Omega$ is convex.
So $42=3\times 14 \in \Omega$ since $\Omega$ is weakly multiplicative.
Then $2\in \Omega,42\in \Omega$ yields $[2,42] \subseteq \Omega$ since $\Omega$ is convex.
More generally, if we define a sequence $(u_k)$ by $u_1=5$ and $u_{k+1}=3u_k-1$ then by induction, every $u_k$ is in $\Omega$, qed.
 UPDATE  (in answer to a comment) Here is how the induction works :
Suppose $u_k \in \Omega$. As $u_k$ is congruent to $2$ modulo $3$, it is coprime
to $3$, so $3u_k \in \Omega$ since $\Omega$ is weakly multiplicative. Then $2$
and $3u_k$ are both in $\Omega$, so $[2,3u_k] \subseteq \Omega$ since $\Omega$ is convex.
In particular, $3u_k-1 \in \Omega$, i.e. $u_{k+1} \in \Omega$. So $\Omega$ contains all the $u_k$ and finally $\Omega$ contains all the positive integers.
