Iranian Mathematical Olympiad

First it is easy to show that the sequence is strictly increasing. Assume that $a_i=a_{i+1}$ for some $i$. Let $j=p-i$ for a large prime $p$. Now $i+j$ is a prime number. So $a_i+a_j$ is prime too. But from $a_i=a_{i+1}$ we get $a_{i+1}+a_j$ and $i+1+j$ are prime numbers. So $i+j$ and $i+j+1$ are two consecutive large prime numbers, contradiction.

Now put $i=j=2^{p-2}$ for a large prime $p$. So the number of divisors of $2a_i=2^{p-1}$ equals $p$. Hence $2a_i$ is of the form $q^{p-1}$ for a prime $q$. Obviously $q$ should be $2$ and therefore we have $a_{2^{p-2}}=2^{p-2}$.

Now we have a strictly increasing sequence of integers with infinitely many fixed points. So, for each $n$, $a_n=n$. $\square$