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Math Help - Polynomial with integer solutions.

  1. #1
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    Polynomial with integer solutions.



    Am I doing it correctly? What would the next step be?
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  2. #2
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    If you let y = x + k, then k must be an integer and x must be an integer solution to x^2 + kx - 12k = 0. Hence, the discriminant of this polynomial, which is k^2 + 48k = k(k + 48), must be a square. Finally, let k = n - 24. Then (n - 24)(n + 24) = n^2 - 24^2 must be a square. Now use your knowledge of pythagorean triples and multiples of them to find solutions:

    (3, 4, 5) \to (24, 32, 40); (18, 24, 30)
    (5, 12, 13) \to (10, 24, 26)
    (7, 24, 25)
    (8, 15, 17) \to (24, 45, 51)
    (9, 40, 41)
    (11, 60, 61)
    (12, 35, 37) \to (24, 70, 74)
    (24, 143, 145)

    Now we need hypotenuses n > 24. From these triples we get the solutions 25, 26, 30, 40, 51, 74, 145.

    This yields the following solutions for k:
    1, 2, 6, 16, 27, 50, 121

    That gives you seven quadratic equations to solve for x, some of which may not yield integer solutions.
    Last edited by icemanfan; March 1st 2010 at 05:52 PM.
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  3. #3
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    Hence, the discriminant of this polynomial must be a square.
    Why?

    Edit: Quadratic formula? Since we want integer solutions, the discriminant must be a perfect square?

    Now use your knowledge of pythagorean triples and multiples of them to find solutions:
    Sorry, but solutions to what? (n-24)(n+24)? or n^2 = 24^2? In either case, where do Pythagorean triples come into play?

    Edit: Pythagorean Theorem. n=24, (n)^2 - 24^2 = k^2 \implies k^2 + 24^2 = n^2. Is that about right?

    Now we need n > 24.
    I'm assuming the answer to this follows from question (2) above.

    Edit: k = n - 24, k > 0, so n > 24. Although, shouldn't this go without saying since n = 24, n^2 = 24^2 + k^2 \implies k = 0, which doesn't make sense?

    Thanks so much for your help.
    Last edited by davismj; March 1st 2010 at 05:38 PM. Reason: Oh I see!
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  4. #4
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    Sorry, that was a hopelessly convoluted method. You are looking for any value of n such that n^2 - 24^2 = a^2 for some integer a.

    I am editing my post to take out values of n that don't work. Where the Pythagorean triples come in: n is the hypotenuse of a right triangle, with one side equal to 24, and the other side equal to a.

    It is much easier to observe that x must be an integer from 1 to 11, and try each of those values and determine whether y ends up being an integer.
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  5. #5
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    One quick question: Is there another way to solve this without the knowledge of pythagorean triples that I don't have?

    Thanks
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  6. #6
    Super Member Bacterius's Avatar
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    Hello !

    You want to solve \frac{1}{x} + \frac{1}{y} = \frac{1}{12} for x and y. This is how I would have done :

    \frac{1}{x} = \frac{1}{12} - \frac{1}{y}

    \frac{1}{x} = \frac{y}{12y} - \frac{12}{12y}

    \frac{1}{x} = \frac{y - 12}{12y}

    x = \frac{12y}{y - 12}

    All you have left is to determine for which values of y do we have an integer x, that is, when (y - 12) | 12y

    Does it make sense ?
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  7. #7
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    Quote Originally Posted by Bacterius View Post
    Hello !

    You want to solve \frac{1}{x} + \frac{1}{y} = \frac{1}{12} for x and y. This is how I would have done :

    \frac{1}{x} = \frac{1}{12} - \frac{1}{y}

    \frac{1}{x} = \frac{y}{12y} - \frac{12}{12y}

    \frac{1}{x} = \frac{y - 12}{12y}

    x = \frac{12y}{y - 12}

    All you have left is to determine for which values of y do we have an integer x, that is, when (y - 12) | 12y

    Does it make sense ?
    Yes, but you copied the problem incorrectly.
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  8. #8
    Super Member Bacterius's Avatar
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    Oh, yeah. Well it doesn't actually change much, just replace the signs
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