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Math Help - Diophantine equation

  1. #1
    Super Member PaulRS's Avatar
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    Diophantine equation

    I have been trying to prove that 12\cdot{x^3}-3=n^2 has no solutions in naturals numbers with x>1

    I first thought of finding a mod in which it makes no sense, but I just couldn't find one (after some attempts).



    Thanks in advance
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  2. #2
    Super Member fardeen_gen's Avatar
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    Are we allowed to use Fermat's Last theorem?

    3\mid n and n > 0 since x > 1.

    Let n = 3m for some integer m > 0.

    Then,
    12x^{3}-3 = 9m^{2}\Rightarrow 4x^{3}-1 = 3m^{2}

    Since the left-hand side is an odd number, m is odd.

    Let m = 2k + 1 for some integer k\geq 0.

    By expanding, we get that
    4x^{3}= 3(2k+1)^{2}= 12k^{2}+12k+4

    Therefore, by dividing by four on both sides,
    x^{3}= 3k^{2}+3k+1\Rightarrow x^{3}+k^{3}= (k+1)^{3}

    If k = 0 , then x = 1 which is impossible since x > 1.

    Otherwise, k > 0 and by Fermat's Last Theorem, there are no positive integer solutions to this equation.

    P.S. I am sure that x^{3}+k^{3}= (k+1)^{3} not having positive integer solutions can be proved using elementary means.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by fardeen_gen View Post
    P.S. I am sure that x^{3}+k^{3}= (k+1)^{3} not having positive integer solutions can be proved using elementary means.
    Fermat's Last Theorem: Fermat's Last Theorem: Proof for n=3
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