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Math Help - Finding solutions

  1. #1
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    Finding solutions

    How would you be able to find the number of solutions on the next equation?

    x^2-4xy+5y^2+2y-4=0

    Given only this, how could you know the number of possible integer pairs of (x,y)? Please show the steps and explain.
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  2. #2
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    Just a clarification for those that want to help me solve this, I will not have a calculator at hand when doing this type of problem so the most important thing is not the answer but rather the logic and procedure you used to get to the answer as that can help me in the future solve other similar problems. Thank you in advance for all your work.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Solaris123 View Post
    How would you be able to find the number of solutions on the next equation?

    x^2-4xy+5y^2+2y-4=0

    Given only this, how could you know the number of possible integer pairs of (x,y)? Please show the steps and explain.
    This equation equals

    (x^2-4xy+4y^2)+(y^2+2y+1)-5=0\implies(x-2y)^2+(y+1)^2=5

    Therefore either

    x-2y=\pm2 and y+1=\pm1

    or

    x-2y=\pm1 and y+1=\pm2

    Solving those equations will get you the (x,y) pairs that are integer solutions.
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  4. #4
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    Ok, that makes sense but before I can say Im comfortable with this, how did you make those squares? I mean, you are right, this makes sense but is there any sort of rule I must follow to find them or I just have to play with values until I make those squares appear?
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  5. #5
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    Quote Originally Posted by Solaris123 View Post
    How would you be able to find the number of solutions on the next equation?

    x^2-4xy+5y^2+2y-4=0

    Given only this, how could you know the number of possible integer pairs of (x,y)? Please show the steps and explain.
    x^2 - 4xy + 5y^2 + 2y - 4 = 0

    You will want to get the equation to be y in terms of x, so move everything without a y to the other side.

    5y^2 + (2 - 4x)y = 4 - x^2.

    Now you will need to complete the square:

    y^2 + \frac{2(1 - 2x)}{5}\,y = \frac{4 - x^2}{5}

    y^2 + \frac{2(1 - 2x)}{5}\,y + \left(\frac{1 - 2x}{5}\right)^2 = \frac{4 - x^2}{5} + \left(\frac{1 - 2x}{5}\right)^2

    \left(y + \frac{1 - 2x}{5}\right)^2 = \frac{(4 - x^2)^2}{25} + \frac{(1 - 2x)^2}{25}

    \left(y + \frac{1 - 2x}{5}\right)^2 = \frac{16 - 8x^2 + x^4 + 1 - 4x + 4x^2}{25}

    \left(y + \frac{1 - 2x}{5}\right)^2 = \frac{x^4 - 4x^2 - 4x + 17}{25}

    y + \frac{1 - 2x}{5} = \pm \frac{\sqrt{x^4 - 4x^2 - 4x + 17}}{5}

    y = \frac{1 - 2x \pm \sqrt{x^4 - 4x^2 - 4x + 17}}{5}.


    Now you need to determine what values of x are going to give you a multiple of 5 in the numerator.
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Solaris123 View Post
    Ok, that makes sense but before I can say Im comfortable with this, how did you make those squares? I mean, you are right, this makes sense but is there any sort of rule I must follow to find them or I just have to play with values until I make those squares appear?
    I hate to rain on your parade, but there's no general algorithm for solving Diophantine equations. (This was Hilbert's Tenth Problem.) You just have to play with each one as best as you can to try to devise a solution. Looking for squares is always a good place to start though.
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  7. #7
    Super Member 11rdc11's Avatar
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    Just wondering is this not an ellipses with a rotation of axes?
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  8. #8
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    Just passed it throught Wolfram Alpha and indeed it's an ellipse with it's two foci slanted, does that make a difference or help solve the problem in any way?
    PS: Anyone know how to use the fact that it is an ellipse to solve it?
    Last edited by Solaris123; March 17th 2010 at 03:13 PM.
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  9. #9
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    Bump
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  10. #10
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Solaris123 View Post
    Just passed it throught Wolfram Alpha and indeed it's an ellipse with it's two foci slanted, does that make a difference or help solve the problem in any way?
    PS: Anyone know how to use the fact that it is an ellipse to solve it?
    The equation would not be anything easier to solve
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