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    I'm trying to find positive rational solutions for the following equation:

    \tan^{-1}x + \tan^{-1}y + \tan^{-1}(\frac{6}{xy}) = \pi

    Obviously any pair of 1, 2, 3 are solutions.

    Clues please because I'm not sure where I'm going with this one.
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    (Original post by Mr M)
    I'm trying to find positive rational solutions for the following equation:

    \tan^{-1}x + \tan^{-1}y + \tan^{-1}(\frac{6}{xy}) = \pi

    Obviously any pair of 1, 2, 3 are solutions.

    Clues please because I'm not sure where I'm going with this one.
    Have you gotten to the stage where you need to solve x^2y+xy^2-6xy+6=0 by taking the tan of both sides from the start? There's symmetry in there, perhaps that helps. You could rearrange this to be in the form of a quadratic in x and then consider the formula to get an answer in terms of y and repeat this for a quadratic in y (and you'll find that these two equations are the same, just with the x and y's swapped around). Although, I'm not sure how I would finish off the problem.
    EDIT: New idea, divide through the above equation by 6xy. May leave you with something much easier to deal with.
    EDIT2: Drawing the graph may also be useful.
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    (Original post by Farhan.Hanif93)
    Have you gotten to the stage where you need to solve x^2y+xy^2-6xy+6=0 by taking the tan of both sides from the start? There's symmetry in there, perhaps that helps. You could rearrange this to be in the form of a quadratic in x and then consider the formula to get an answer in terms of y and repeat this for a quadratic in y (and you'll find that these two equations are the same, just with the x and y's swapped around). Although, I'm not sure how I would finish off the problem.
    I did get to there and tried to look at what I needed under the square root so that the answers would be rational. I couldn't get that to work. The answers do lie between 0.9358 and 3.3055 but that doesn't really help either.
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    (Original post by Mr M)
    I did get to there and tried to look at what I needed under the square root so that the answers would be rational. I couldn't get that to work. The answers do lie between 0.9358 and 3.3055 but that doesn't really help either.
    I've gotten down to this:
    x+y+\frac{6}{xy}=6
    I'm not sure how to get any further. If you do figure it out, could I ask you to post your solution up on here as I think I'd be interested to see it? Would have been so much easier if x and y were just integers...
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    (Original post by Farhan.Hanif93)
    I've gotten down to this:
    x+y+\frac{6}{xy}=6
    I'm not sure how to get any further. If you do figure it out, could I ask you to post your solution up on here as I think I'd be interested to see it? Would have been so much easier if x and y were just integers...
    Yes, how inconvenient that the number of rationals is not finite (and small)!
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    (Original post by Mr M)
    Yes, how inconvenient that the number of rationals is not finite (and small)!
    Indeed, I think I'll have to leave this for someone with a few more brain cells than me to round off.
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    (Original post by Farhan.Hanif93)
    Have you gotten to the stage where you need to solve x^2y+xy^2-6xy+6=0 by taking the tan of both sides from the start? There's symmetry in there, perhaps that helps. You could rearrange this to be in the form of a quadratic in x and then consider the formula to get an answer in terms of y and repeat this for a quadratic in y (and you'll find that these two equations are the same, just with the x and y's swapped around). Although, I'm not sure how I would finish off the problem.
    EDIT: New idea, divide through the above equation by 6xy. May leave you with something much easier to deal with.
    EDIT2: Drawing the graph may also be useful.
    That simplifies to

     x^2 + (y-6)x + \frac{6}{y} = 0

    After that wouldn't you just put some numbers to produce a pair of solutions i.e putting y = 2 gives x = 1.

    which is a valid pair of solution.
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    (Original post by jasbirsingh)
    That simplifies to

     x^2 + (y-6)x + \frac{6}{y} = 0

    After that wouldn't you just put some numbers to produce a pair of solutions i.e putting y = 2 gives x = 1.

    which is a valid pair of solution.
    Yes but how would you find ALL of the valid rational solutions? That's not quite so easy. I found all the integer solutions instantly. Finding all of the rational solutions is far more difficult than that.
    It appears to be easier to work with x+y+\frac{6}{xy}=6 than what you have up there.
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    (Original post by Farhan.Hanif93)
    Yes but how would you find ALL of the valid rational solutions? That's not quite so easy. I found all the integer solutions instantly. Finding all of the rational solutions is far more difficult than that.
    It appears to be easier to work with x+y+\frac{6}{xy}=6 than what you have up there.
    Oops sorry I didn't read the question properly.
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    (Original post by Farhan.Hanif93)
    Yes but how would you find ALL of the valid rational solutions? That's not quite so easy. I found all the integer solutions instantly. Finding all of the rational solutions is far more difficult than that.
    It appears to be easier to work with x+y+\frac{6}{xy}=6 than what you have up there.
    Actually, I haven't been asked to find ALL of them. One more will do but I can't even manage that.

    :sad:
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    By computer search:
    When y = 8, you get x^2 + 2x + 3/4 = 0, so x = -1/2 or x=-3/2 both work.

    Also y = -3/2, y = 25/21, y= 49/15, y = 54/35, -361/68, 867/76. (Haven't verified whether any of these work, although y = -3/2 is a symmetry pair to the y=8 case so it will).
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    (Original post by DFranklin)
    By computer search:
    When y = 8, you get x^2 + 2x + 3/4 = 0, so x = -1/2 or x=-3/2 both work.

    Also y = -3/2, y = 25/21, y= 49/15, y = 54/35, -361/68, 867/76. (Haven't verified whether any of these work, although y = -3/2 is a symmetry pair to the y=8 case so it will).
    I believe that Mr M asked for the positive rational solutions for x and y.
    EDIT: (x,y)=(\frac{49}{15},\frac{25}{2  1}) is a suggestion of yours that works.
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    Guessing somewhat about which solutions will 'pair', I suspect you'll find y=49/15, x = 54/35 will be such a solution.

    (Haven't time to check right now).
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    (Original post by DFranklin)
    Guessing somewhat about which solutions will 'pair', I suspect you'll find y=49/15, x = 54/35 will be such a solution.

    (Haven't time to check right now).
    Thanks. That has moved me forward anyway.
 
 
 
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