$ ab-1|a^2+ab+b^2 $ I hava a number theory problem. I think on it yestarday night and today, afternoon.
The problem :
$ a,b $ are two natural numbers such that : $ ab>1 $
how many pairs $ (a,b) $ is there such that : $ ab-1|a^2+ab+b^2 $
we have $ ab-1|(a^2+ab+b^2)+(ab-1) = (a+b)^2 - 1^2 = (a+b+1)(a+b-1) $
so $ ab-1|(a+b+1)(a+b-1) $
can you help me to complete this?!
i need a hint!
please don't write a compelete solution : )

$a + b + 1$ and $a + b - 1$ are either relatively prime or have $2$ as a common divisor. In the first case, we must have $ab - 1$ divides $a + b + 1$ or $ab - 1$ divides $a + b - 1$. In the second case, there are a few more options to consider.
