$ ab-1|a^2+ab+b^2 $

Question

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 : )

Answer 1

Hint: Look up vieta jumping. This is a classic "olympiad" number theory technique although it is not very enlightening.

Answer 2

$a + b + 1$ 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.