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Math Help - Simple Diophantine Equation

  1. #1
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    Simple Diophantine Equation

    Given a prime  p , show  \displaystyle x^3+py^3+p^2z^3=0 has no nontrivial integer solutions.
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  2. #2
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    Proof by infinite descent:

    Suppose there axists a solution 0\ne (x,y,z). Then p\mid x^3 since p\mid (p^3+p^2z^3).

    Therefore we also have p\mid x \Rightarrow x = ap.

    Thus, after substitution and division by  p we get y^3+pz^3+p^2a^3 = 0.

    An analogous argument shows that p \mid y and then p\mid z and we get

    a new solution (a,b,c) = (\frac{x}{p},\frac{y}{p},\frac{z}{p}).

    We can proceed with this argument infinitely many times, which is of course absurd.


    Note: This problem can be generalized to the statement that there are no

    solutions 0\ne (x_1,x_2,...,x_n) \in \mathbb{Z}^n to \sum_{k=1}^{n} p^kx_k^n = 0

    (EDIT) where p is squarefree.
    Last edited by Unbeatable0; August 17th 2010 at 02:47 AM.
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  3. #3
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    First part, to prove divisibility by the prime p for x, y, and z . From x^3+py^3+p^2z^3=0 it's clear that p|x and then p|x^3 . Dividing the equation by p gives x^3/p+y^3+pz^3=0 and since p still divides x^3/p it follows that p|y , also p|y^3 . Dividing again by p gives x^3/p^2+y^3/p+z^3=0 , where x^3/p^2 and y^3/p are divisible by p so p|z .

    x^3, y^3, and z^3 are of the forms Ap^3, Bp^3, and Cp^3 , respectively.Also, A, B,and C are relatively prime to p . Therefore, p^{3s}A+p^{3t+1}B+p^{3u+2}C=0 .

    p^{3s}\neq p^{3t+1}\neq p^{3u+2} . If p^{3s} is the smallest of these, then after division we have A+p^{3T+1}B+p^{3U+2}C=0 . But then p|A ; since (p,A)=1 and A>1 it's contradiction.
    Similarly for the cases where p^{3t+1} or p^{3u+2} are the smallest powers.
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