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Find only the integer zeros?

  1. Aug 27, 2011 #1
    Hello every one, please help me finding the integer zeros for this equation

    -15x^3+45x^2y-15xy^2+y^3+90x^2-210xy+40y^2-120x+184y = 0

    I know that the solution should used by diophantine equation but i don't know
    ho to solve it, Please help :(
     
  2. jcsd
  3. Aug 27, 2011 #2

    Hurkyl

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    I don't think it's especially easy to do so -- barring something unusual (and I don't see anything unusual), (total) degree 3 polynomial equations in two variables define a type of plane curve called an "elliptic curve", and I am vaguely under the impression that the problem of finding integer points (a.k.a. integral points) on elliptic curves is fairly difficult.
     
  4. Aug 27, 2011 #3

    micromass

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    These kinds of problems can be very difficult. A very good book written on the subject is "Rational points on Elliptic Curves" by Silverman and Tate.

    The first thing that we should do here is to reduce this equation to the Weierstrass normal form. That is, we would like to express the equation as

    [tex]y^2=x^3+ax^2+bx+c[/tex]

    This can be done by a bit projective geometry. We begin by taking a rational point on the original cubic (for example (0,0)). We let Z=0 to be the tangent line to (0,0). Now, the tangent line will intersect our cubic in another rational point. Let X=0 be the tangent line to this other point and let Y=0 be an arbitrary line through (0,0) (which is not the tangent line).

    Now, if we choose axes in this manner, and if we apply the projective transformation [itex]x=X/Z[/itex]and [itex]y=Y/Z[/itex] then we will get an equation of the form

    [tex]xy^2+(ax+b)y=cx^2+dx+e[/tex]

    Try to work this out first, then we can go further in the reducing.
     
  5. Aug 27, 2011 #4
    Thanks,,

    the global minimal Weierstras is
    y^2+xy+y=x^3-x^2-62705x+5793679.
     
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