# Thread: Extremely difficult IMO problem

1. ## 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 $\frac{a^2}{2ab^2-b^3+1}$ is a positive integer."

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

2. Originally Posted by Jameson
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 $\frac{a^2}{2ab^2-b^3+1}$ is a positive integer."

The only assertion I can make is that $2ab^2-b^3+1>0$.
Are a, b integers as well? If so, then another criterion comes from the rational roots test:

Let n be a positive integer, then let
$\frac{a^2}{2ab^2-b^3+1} = n$

So $nb^3-2nab^2+(a^2-n)=0$

Thus we know that the only rational b values that satisfy this equation are of the form:
$b = \frac{factor \, of \, a^2-n}{factor \, of \, n}$

Since b must be an integer, the only possible values for b are those for which factors of n are factors of a^2-n.

Just a thought.

-Dan

3. Hmmm...

I just found a non-integer pair $\left ( \frac{1 \pm \sqrt{15}}{4}, \, \frac{1}{2} \right )$ that seems to work.

I guess they DON'T need to be integers...

-Dan

4. (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:

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

as long as the discriminant is real.

-Dan

5. Could you show your work for your last post? I appreciate your insight. The question asks for all pairs, and I'm getting from your work that there could be infinite. How would I go about first finding the number of pairs?

6. Originally Posted by Jameson
Could you show your work for your last post? I appreciate your insight. The question asks for all pairs, and I'm getting from your work that there could be infinite. How would I go about first finding the number of pairs?
$\frac{a^2}{2ab^2-b^3+1}=n$
Thus,
$a^2=2ab^2n-b^3n+n$

7. Originally Posted by ThePerfectHacker
topsquark (the physicist)
Do I detect a note of disdain here? (I think he's just jealous. ) And hey, even the workers at Friendly's have enough respect to call me "The Professor!"

-Dan

8. by the way the actual question is:
Determine all pairs of positive integers $(a,b)$ such that
$\frac{a^2}{2ab^2-b^3+1}$
is a positive integer.

so a and b must be positive integers.

here are a few examples of positive integers which work:
a = 126, b = 4
a = 7, b = 2
a = 116149892142279364320, b = 123456

9. ## Are there any more pairs except these?

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:

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

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

10. The more trivial ones.
(a,b)=(k,2k)
(a,b)=(2k,1)

11. Originally Posted by ThePerfectHacker
The more trivial ones.
(a,b)=(k,2k)
(a,b)=(2k,1)
The method I used show that there should be two values of a for each b
i.e. if b=2k either a = 8*(k^4) - k or a=k(a=k is not trivial)
The trivial one is (2k,1)
Can b be a odd no. except 1? I think no. What do you think?

12. We have,
$\frac{a^2}{2ab^2-b^3+1}=k\geq 1$
Thus,
$a^2-2ab^2k+b^3k-k=0$
Let us find under which conditions this quadradic has a unique solution, that is when,
$4b^4k^2-4b^3k+4k=0$
Thus,
$b^3k-b^2+1=0$
Thus,
$k=\frac{b^2-1}{b^3}$
Is not an integer because the numerator factors into $(b-1)(b+1)$ and $\gcd(b,b+1)=1$ therefore the only case is $b=1$, which leads to our pairs $(2k,1)$.
Otherwise what I have showed is that if this equation has two solutions and one is a positive integer then the other must be also because it follows by Viete's theorem.