Find all odd positive integers greater than 1 such that for any relatively prime divisors a and b of n, a+b-1 is also a divisor of n 
Somewhy I think that all such numbers are an empty set cuz a+b-1 is also odd, hence U can find some d: $d<n$, $d<a+b-1$, which is also odd, and relatively prime to $(a+b-1)=f$, and then again, f+d-1 must be a divisor of n, and so on, and so on, but at one step $f*+d*-1$ will be greater than n, hence contradiction..any thoughts?

Write $M = p^n m$, where $p$ is the smallest prime factor of $M$, and $(p^n,m) = 1$. Since $p^n$ is coprime to $m$, it is true that $p+m - 1 | M$. However, since $p+m-1$ is co-prime to $m$, as $p-1$ is co prime to $m$, because $m$ contains only prime factors greater than $p$, it follows that $p+m-1 \stackrel{(1)}{=} p^k$  for some $k$. 
Now, again, $p^k + m - 1 = 2p^k - p$ is a divisor of $M$, which implies that $2p^{k-1} - 1$ is a divisor of $M$. However, since it is co prime to $p^k$, it will have to divide $m$. However, this implies that:
$$
2p^{k-1} - 1 | p (2p^{k-1}-1) - 2m = p-2
$$ by $(1)$.
This funny occasion occurs precisely when $k = 1$, since otherwise $2p^{k-1}-1 > p - 2$. If that happens, then of course $m=1$, and then $M = p^k$, where $p$ is a prime. You can check that it indeed holds true for prime powers, and hence the property holds true precisely for prime powers.
