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Math Help - Satisfying the equation

  1. #1
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    Satisfying the equation

    Find all pairs of positive integers x and y that satisfy the equation.
    1/x - 1/y = 1/2005

    (y-x)/xy=1/2005
    2005(y-x) = xy

    Is this a good method?
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  2. #2
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    So far so good.
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  3. #3
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    But I couldn't find a way to continue. Do you have any hints?
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  4. #4
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    Get everything over on to one side:
    xy - 2005(y-x) = 0, that is
    xy + 2005x - 2005y
    Try to factorise: (x + ?)(y + ?)
    (x-2005)(y+2005) = 2005^2
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  5. #5
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    I tried to continue and put factors in, but they don't seem to work.
    Can you explain a bit further if you can?

    Or are there other easier methods?

    Thanks!
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  6. #6
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    ok

    So you have (x-2005) * (y+2005) = 2005^2. The key now is to see that the equation tells you that (x-2005) and (y+2005) are factors of 2005^2. Moreover, when multiplied together, they give 2005^2.

    2005^2 in prime factors factorization : 2005^2 = 5^2 * 401^2 (I think 401 is prime)

    So you don't have many choices (you choose from those factors some for (x-2005) and keep the rest for (y+2005)) :

    1) (x-2005) = 1 and so, (y+2005)=5^2 * 401^2. We get : x=2006 and y=4018020
    2) (x-2005) = 5 and so, (y+2005)=5 * 401^2. We get : x=2010 and y=802000
    3) (x-2005) = 5^2 and so, (y+2005)=401^2. We get : x=2030 and y=158796
    4) (x-2005) = 401 and so, (y+2005)=5^2 * 401. We get : x=2406 and y=8020
    5) (x-2005) = 5*401 and so, (y+2005)=5*401. We get : x=4010 and y=0(we reject it because y=0 and so is not positive - if the teacher wanted the y=0, he/she would have said give me x and y non-negative. So positive means >0).
    6) (x-2005) = 401^2 and so, (y+2005)=5^2. We get : x=162806 and y=-1980 (we reject it because y<0)

    And so we must regect all other combos for x because it gives (y+2005) <2005 which gives y<0 (not wanted).

    So the main work was the factorization done by "rgep". After that, you look for factors.

    For a very similar example see : ryanchow (need help!!!) in the urgent math homework section
    Last edited by hemza; September 10th 2005 at 07:45 AM.
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  7. #7
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    Thanks for your explanation. I've tried this method before, but when you put the x and y into the original equation, they don't work.
    Last edited by ardnas; September 10th 2005 at 06:43 PM.
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  8. #8
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    Sorry

    Hi,

    I am sorry. There was a little sign mistake in rgep factorization. Here is what he did:

    "Get everything over on to one side:
    xy - 2005(y-x) = 0, that is
    xy + 2005x - 2005y
    Try to factorise: (x + ?)(y + ?)
    (x-2005)(y+2005) = 2005^2 "

    But it should be :
    2005(y-x) = xy (from your calculations)
    2005y-2005x-xy=0
    We add 2005^2 for both sides: 2005^2 +2005y- 2005x-xy=2005^2
    We factorize : 2005(2005+y)-x(2005+y)=2005^2
    We factorize again : (2005-x)(2005+y)=2005^2 (you see there is a little sign change)

    I you use after that the same method I used on my previous reply for finding x and y you'll get the answers wanted.

    Sorry, I should have checked the factorization before.

    If you have difficulty finding the positive integers x and y satisfying the equation, respond to this "reply" and I'll help you with it.
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  9. #9
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    Thanks so much!
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