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 is a positive integer."

The only assertion I can make is that .

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- April 6th 2006, 11:10 AMJamesonExtremely 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 is a positive integer."

The only assertion I can make is that . - April 6th 2006, 01:43 PMtopsquarkQuote:

Originally Posted by**Jameson**

Let n be a positive integer, then let

So

Thus we know that the only rational b values that satisfy this equation are of the form:

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 - April 6th 2006, 02:02 PMtopsquark
Hmmm...

I just found a non-integer pair that seems to work.

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

-Dan - April 6th 2006, 02:08 PMtopsquark
(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:

as long as the discriminant is real.

-Dan - April 6th 2006, 03:54 PMJameson
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?

- April 6th 2006, 04:07 PMThePerfectHackerQuote:

Originally Posted by**Jameson**

Thus,

Then, he just used quadradic formula. - April 7th 2006, 06:25 AMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

-Dan - April 8th 2006, 02:05 AMAradesh
by the way the actual question is:

Determine all pairs of positive integers such that

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 - June 6th 2006, 11:51 PMmalaygoelAre there any more pairs except these?Quote:

Originally Posted by**topsquark**

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. - June 7th 2006, 11:30 AMThePerfectHacker
The more trivial ones.

(a,b)=(k,2k)

(a,b)=(2k,1) - June 7th 2006, 05:42 PMmalaygoelQuote:

Originally Posted by**ThePerfectHacker**

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? - June 7th 2006, 06:30 PMThePerfectHacker
We have,

Thus,

Let us find under which conditions this quadradic has a unique solution, that is when,

Thus,

Thus,

Is not an integer because the numerator factors into and therefore the only case is , which leads to our pairs .

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.