Extremely difficult IMO problem

This problem is from the 2003 IMO, the third level of the American Mathematics Competition. I do not even know how to approach this.

"Determine all paris (a,b) such that $\displaystyle \frac{a^2}{2ab^2-b^3+1}$ is a positive integer."

The only assertion I can make is that $\displaystyle 2ab^2-b^3+1>0$.

Are there any more pairs except these?

Quote:

Originally Posted by **topsquark**

(Shrugs) I guess by letting n be a positive integer and otherwise letting n and b be arbitrary we can always find two "a" values for each (b, n) pair:

$\displaystyle a = nb^2 \pm \sqrt{n^2b^4-nb^3+n}$

as long as the discriminant is real.

-Dan

I visited this thread yesterday and the question sounded interesting to me.

Continuing the quoted work, I tried to find the integer values(of b and n) for which discriminant is real and a perfect square(so that a is an integer), and hence the answer to the question.

I found that **a= 8*(k^4) - k**

b= 2k

where k is any positive integer gives the pairs pair satisfying the conditions in the question.

e.g

k=1, b=2, a=7

k=2, b=4, a=126

k=3, b=6, a=645

...................k is any positive real number

Are there any more pairs?

Please verify the above answers.

If you want to know how I arrived at the generalization, tell me.