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A diophantine equation

  1. May 28, 2006 #1
    I think I posted this in the wrong forum before. Let's try again.

    I need to prove that the equation

    [itex]x^3 + y^3 = 3z^3[/itex]

    has no integer solutions. I can do it easily for all cases except where z has a factor of 3, in which case I don't know what to do.

    I am assuming the 3 in front of the z term is supposed to make this easier somehow than the same equation without it, but I'm failing to see the simplification that this allows.

    Anyone know?
  2. jcsd
  3. May 29, 2006 #2
    Well, one thing is sure 3 divides (X+Y), and it follows that 3 divides X^2-XY+Y^2, by modulo arithmetic, but only once. Proof:

    Y==-X Mod 9 implies X^2-X(-X)+(-X)^2==3X^2 Mod 9. But 3 divides X implies 3 divides Y and Z. So we can remove such a factor.

    So that we have X+Y =(3a)^3, X^2-XY+Y^2=3b^3. Z=3ab.

    Well, maybe that helps....
  4. May 29, 2006 #3
    The only problem is if z has a factor of 3. We can then write:

    [itex]x^3 + y^3 = 3^4z'^3[/itex]

    If we then set:


    Then we have 4 factors of 3 to divide up between p and q. It is easy to show that all options are impossible except one. 27 divides p and 3 divides q.

    All cases except this one are easy. This is what I am looking for some insight on (or a completely different approach that bypasses the need to divide it up like this).
  5. May 30, 2006 #4
    Here I am providing a very simple solution(only for natural numbers). If you find any mistake, tell me.
    The solution goes like this:
    3z^3 = x^3 +y^3 =(x+y)(x^2 – xy + y^2)
    Now two things to note here:
    (1)z>3(you can check it yourself)
    (2)The first factor on the right hand side of the equation(x+y) is less than the second(x^2 – xy + y^2)(I can prove it to you if you want)

    Now I will factorize 3z^3 into two factors with smaller first:
    (1) 1,3z^3
    (2) 3,z^3
    (3) 3z,z^2
    (4) z,3z^2
    (Are there any more factors?)
    You can easily see first and second case are not applicable
    (3) x+y=3z
    (x^2 – xy + y^2)=z^2
    Eliminating z we get
    {(x+y)/3}^2 = (x^2 – xy + y^2)
    (x^2 + 2xy + y^2) = 9(x^2 – xy + y^2)
    8(x^2 – 2xy + y^2) = -5xy, .. … a contradiction

    (4) x+y=z
    (x^2 – xy + y^2)=3z^2 =3 (x+y)^2 =3(x^2 + 2xy + y^2)
    2(x^2 – 2xy + y^2)= -9xy,…..a contradiction

    all cases exhausted
  6. May 31, 2006 #5
    You can't set composite factors equal like that.
  7. Jun 1, 2006 #6
    you can substitute
    it will simplify your calculations
    try it please
  8. Jun 1, 2006 #7
    check out the chapter (if you have not yet done it) on "some diophantine equations" in Hardy's "An Introduction to the Theory of Numbers". there is a proof of exactly what you need, gonzo.
  9. Jun 1, 2006 #8
    Thanks, but I don't have that book and I doubt it's in my library. I found a solution anyway using infinite descent.
  10. Jun 7, 2006 #9
    Ribenboim points out that Lagrange was the first to show no solutions to [tex]X^3+Y^3=3Z^3[/tex], and that much work has been done on the equation: [tex]X^3+Y^3=aZ^3[/tex], though he did not give details.

    IA Barnett, Elements of Number Theory points out that no solution for [tex]X^P+Y^P=PZ^P[/tex] exists, P>2, unless P divides Z. (This is relatively easy to show) HOWEVER, nothing is said about the case where P DOES DIVIDE Z, and I am left wondering if this matter is largely unknown.
    Last edited: Jun 8, 2006
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