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Find all ordered pair of integers for a diophantine equation

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data
    find all orderde pairs of integers (x,y) such that x^2+y^2=4x+2y


    2. Relevant equations



    3. The attempt at a solution
    rearrange to--> x^2=4x+2y-y^2
    because x and y can only be integers, y(2-y) must be divisible by x
    so y(2-y)>=x
    y(2-y)=x(x-4)
    x(x-4)>=x
    x-4>=1
    x>=5

    however, if i test with the ordered pair (3, 3) the diophantine equation is true
    x^2+y^2=4x+2y
    9+9=12+6
    18=18

    help me please!!!! thanks
     
  2. jcsd
  3. Sep 16, 2009 #2

    Dick

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    Complete the square in both x and y. I.e. write it as (x-a)^2+(y-b)^2=c. Figure out a, b and c. Now try and solve it again. You'll find it's pretty easy.
     
  4. Sep 16, 2009 #3

    ehild

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    y(y-2) does not need to be greater than x. It can be equal, and can be negative.
    You have the equation in form y(2-y)=x(x-4). At what condition are both sides positive, 0, and negative?

    ehild
     
  5. Sep 19, 2009 #4
    @dick

    if i complete the square i get (x-2)^2+(y-sqrt(2))^2=6

    i have no idea what to do from there

    @ehild

    x(x-4)=0 when y=0 or y=2
    y(2-y)=0 when x=0 or x=4

    so (0, 0) is one pair

    y(2-y)=x(x-4)

    LS is +ve when 2>y>0
    RS is +ve when x>4

    LS is -ve when y>2 or y<0
    RS is -ve when x<4

    there are no integer solutions when ls and rs are +ve
     
    Last edited: Sep 19, 2009
  6. Sep 19, 2009 #5

    Dick

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    When I complete the squares I get (x-2)^2+(y-1)^2=5. How did you get the sqrt(2) in there?
     
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