Prove that for any integer $n>1$, there exists a set of $n$ positive integers such that, for any two numbers among them (say $a$ and $b$), $a-b$ divides $a+b$
I have come up with three strategies to tackle this problem:
(i) Try to construct a set satisfying the conditions
(iii) Try to prove it by contradiction. (Which is, I think very difficult to do so)
I have tried smaller examples hoping to find a pattern. Tried arithmetic, geometric series, but no luck. It is very hard to even come up with an example for $n=5$. We can make some simple observations like $(n,n+1)$ and $(n,n+2)$ always work. But the thing with this problem, that is making it difficult, is the rule should be followed by every two numbers in the set.
Induction definitely fails, fix any number $a$, then condition $a-x|a+x$ can also be written as $a-x|2a$. Which means there are only finitely many values of $x$ which satisfies the condition. So, we can't rely on induction
I am not sure, how can we go about using (ii)? Or is there any other method?