Find the sequence {$a_n$} so that $a_i + a_j$ has the same amount of dividers as i+j, for every positive integer i and j$ Find all rising sequences positive integers $a_1,a_2,...$ for which holds tgat for every two positive integers i and j the number i+j has the same number of dividers as $a_i + a_j$.
I got that if i=j then $a_i$ and i must havr the same amount of dividers.

Write $\tau(i)$ for the number of divisors of $i$. If $n = p - 2$, where $p$ is a prime, we have
$$
\tau(2^n + 2^n) = \tau(2^{n+1}) = n + 2 = p = \tau(a_{2^n}+a_{2^n}) = \tau(2a_{2^n})
$$
But $\tau(2i)$ is not prime unless $i$ is a power of $2$, so we must have $a_{2^n} = 2^n$ to get $\tau(2a_{2^n}) = p$. So $a_i$ is a strictly increasing sequence of positive integers such that there are arbitrarily large $m$ with $a_m = m$. This implies $a_i = i$ for all $i$.
