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 .
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 .
Are a, b integers as well? If so, then another criterion comes from the rational roots test: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
(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
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
I visited this thread yesterday and the question sounded interesting to me.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.
The method I used show that there should be two values of a for each bOriginally 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?
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.