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Quadratic equation

  1. Jun 19, 2013 #1
    1. The problem statement, all variables and given/known data
    The total number of integers that satisfy the equation x2-4xy+5y2+2y-4=0
     
  2. jcsd
  3. Jun 19, 2013 #2

    haruspex

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    If you can write the variable portion as a sum of squares you'll be able to limit the possibilities considerably.
     
  4. Jun 19, 2013 #3

    lurflurf

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    Yes completing the square is helpful. We can also note the bound
    -3<=y<=2
    So we know at most 12 points need to be checked.
    also
    a^2+b^2
    limits the solutions
     
  5. Jun 19, 2013 #4
    How do you note the bound?How should I complete the square with the 4xy?
     
  6. Jun 19, 2013 #5

    HallsofIvy

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    To complete the square in [tex]x^2+ ax[/tex], divide the coefficient of x, a, by 2 and add the square of that:
    (a/2)^2= a^2/4. Here "a" is 4y. What is a^2/4?
     
  7. Jun 19, 2013 #6

    lurflurf

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    The bound is obvious after the squares are completed. There are a few forms the most useful is

    (x+ay)^2+(y+b)^2+c=0
    expanding
    x^2+2axy+(a^2+1)y^2+2by+(c+b^2)=0
    matching with
    x^2-4xy+5y^2+2y-4=0
    gives
    2a=-4
    a^2+1=5
    2b=2
    c+b^2=-4
     
  8. Jun 20, 2013 #7
    Yea I got that but I still do not know how to continue.Forgive my ignorance
     
  9. Jun 20, 2013 #8

    lurflurf

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    so you know a,b,c
    (x+ay)^2+(y+b)^2+c=0
    c<0
    (y+b)^2<-c
    -b-sqrt(-c)<=y<=-b+sqrt(-c)
    is the desired bound on y
    list out possibilities
    y=-3,-2,-1,0,1,2
    since (x+ay)^2+(y+b)^2+c=0
    what values of a^2+b^2=5
    are lattice points (integers)?
    let
    a=(x+ay)
    b=(y+b)
    complete solution
     
  10. Jun 20, 2013 #9
    Let me see,..why do you say (y+b)^2<=-c
     
  11. Jun 20, 2013 #10

    lurflurf

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    (x+ay)^2+(y+b)^2+c=0
    (y+b)^2=-(x+ay)^2-c
    0<=(x+ay)^2
    (y+b)^2=-c
    if that is a bit obtuse consider
    1+4=5
    notice 1,4,5=>0
    see that we can conclude
    1<5
    in general if
    a+b=c
    a,b,c=>0
    b<=c
     
  12. Jun 20, 2013 #11
    Wow nice,...I guess I will take some time to go over the whole equation again.Thanks for the assistance
     
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