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Math Help - Extremely difficult IMO problem

  1. #1
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    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.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote 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
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  3. #3
    Forum Admin topsquark's Avatar
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    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
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  4. #4
    Forum Admin topsquark's Avatar
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    (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
    Last edited by topsquark; April 6th 2006 at 02:08 PM. Reason: Oops! (I'm doing that a lot today! :) )
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  5. #5
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    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?
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    Quote 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?
    topsquark (the physicist) had,
    \frac{a^2}{2ab^2-b^3+1}=n
    Thus,
    a^2=2ab^2n-b^3n+n
    Then, he just used quadradic formula.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote 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
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  8. #8
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    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
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  9. #9
    Super Member malaygoel's Avatar
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    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:

    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.
    Last edited by malaygoel; June 7th 2006 at 06:02 AM.
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  10. #10
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    The more trivial ones.
    (a,b)=(k,2k)
    (a,b)=(2k,1)
    Last edited by ThePerfectHacker; June 7th 2006 at 01:26 PM.
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  11. #11
    Super Member malaygoel's Avatar
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    Quote 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?
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  12. #12
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    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.
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