here's the problem: Find all odd integers $n$ greater than $1$ such that for any relatively prime divisors $a,b$ of $n$ the number $a+b-1$ is also a divisor of $n$.
And this is my proof (which I believe is incorrect):
Clearly any prime power works, as the only relatively prime divisors are one and the number.
Now, for any integer that isn't a prime power, we can clearly find two non-consecutive relatively prime divisors $a,b$. It's clear that $a+b-1|n$. Since $(a+b-1,a) = 1$ we must clearly have $a+2b-1|n$. We proceed by induction. The base case holds, assuming that $ka+b-1|n$ we clearly have $(ka+b-1,a) = 1$ thus $a(k+1)+b-1|n$, completing our induction. This would mean that $n$ has infinitely many divisors, and so the only solutions are the prime powers.
Could you please tell me what's wrong, and if you have time to, how to correct it? Thanks!