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Diophantine Equations

  1. Mar 11, 2006 #1
    Does anyone know of any resources on the web (or if you prefer, provide me directly with assistance) which will help me understand how to solve equations of the form

    [tex]ax^3+by^3=c[/tex]

    I believe they are third-order Diophantine equations.
     
  2. jcsd
  3. Mar 12, 2006 #2
    Perhaps I should be a little more specific.

    How would you go about proving that a third order Diophantine equation has no solutions?
     
  4. Mar 14, 2006 #3
    I decided to have a go and here is how I went. Let me know if I made any mistakes

    Consider the Diophantine equation

    [tex]x^3+117y^3=5[/tex]

    Choose mod 5.

    The equation tells us that [itex]5|x^3 \Rightarrow 5^3|x^3 \Rightarrow x = 5X[/itex]. Therefore

    [tex]125\cdot X^3 +117y^3 = 5[/tex]

    By the same procedure as above, this equation tells us that [itex]5|y^3 \Rightarrow y = 5Y[/itex]. Therefore

    [tex]125\cdot X^3 + 117\cdot 5Y^3 = 5[/tex]

    Divide through by 5 and we have

    [tex]25\cdot X^3 + 117\cdot Y^3 = 1[/tex]

    This equation now tells me that

    [tex]25X^3 \equiv 1(\mod 5)[/tex]

    But [itex]25\equiv 0 (\mod 5)[/itex]. Hence there is no such [itex]0,1,2,3,4[/itex] such that [itex]X^3\equiv 1(\mod 5)[/itex].

    However, if we had checked Y first we would have found

    [tex]117Y^3 \equiv 1(\mod 5)[/tex]

    which implies that

    [tex]Y^3 \equiv 2(\mod 5)[/tex]

    since [itex]117\equiv 2 (\mod 5)[/itex]. And since if we let [itex]Y = 3[/itex] then

    [tex]Y^3 = 3^3 = 27 \equiv 2(\mod 5)[/tex]

    then this tells us that there is a solution. But since it failed for X there is no solution.
     
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