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Math Help - Prove x^6+7y^3=2 has no solution

  1. #1
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    Prove x^6+7y^3=2 has no solution

    Hi, I have the following problem in a course titled "Introduction to Abstract Algebra"... I am not quite sure if I got the question right though because it looks so easy.

    Thanks for your help!

    "Prove that the equation

    x^6+7y^3=2

    has no solution with x and y being in Z (set of integers)"

    I think that is it because y=(2-x^6)^{1/3}/(7)^{1/3}...but I don't know how to explain it :-/
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  2. #2
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    Quote Originally Posted by mbmstudent View Post
    Hi, I have the following problem in a course titled "Introduction to Abstract Algebra"... I am not quite sure if I got the question right though because it looks so easy.

    Thanks for your help!

    "Prove that the equation

    x^6+7y^3=2

    has no solution with x and y being in Z (set of integers)"
    If you know a bit of modular arithmetic then the quickest way to see this is to notice that x^6\equiv1\pmod7, so you can't have x^6\equiv2\pmod7.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    If you know a bit of modular arithmetic then the quickest way to see this is to notice that x^6\equiv1\pmod7, so you can't have x^6\equiv2\pmod7.
    Ah... simple yet elegant, I like it.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Opalg View Post
    If you know a bit of modular arithmetic then the quickest way to see this is to notice that x^6\equiv1\pmod7, so you can't have x^6\equiv2\pmod7.
    I agree that this is a very nice little piece of work. However, if I may, how did you know that x^6\equiv1\pmod7?

    -Dan
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    Forum Admin topsquark's Avatar
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    Ah! Fermat's Little Theorem. I got it now.

    -Dan
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  6. #6
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    Quote Originally Posted by Opalg View Post
    If you know a bit of modular arithmetic then the quickest way to see this is to notice that x^6\equiv1\pmod7, so you can't have x^6\equiv2\pmod7.
    We must also mention the case: x^6\equiv0\pmod7 (because it's a condition in Fermat's Little Theorem )

    Anyway , nice solution .
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