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Math Help - diophantine equation

  1. #1
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    diophantine equation

    I have to show that the Diophantine eqn.

    x^2 + 10y^2 = 2z^2 has no solutions

    first i let x,y,z be a solution

    then i got x^2 is congruent to 0 (mod 2)

    but i don't know how this helps or any other way to approach it.
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  2. #2
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    Did you try Fermat's Infinite Descent ?
    mmm , sorry after trying it , i don't think it's a good way to treat this exercice .
    But we can also see that this equation is equivalent to this one ,
    (because , as you said , x˛ is congruent to 0 (mod 2) so x is also congruent to 0 (mod 2) , then , x=2x' ...)
    I also think that using inequalities is a good way ..... (i'll give it a try later )

    P.S: Do we have to solve it in |N or Z ?
    Last edited by Tarask; February 28th 2011 at 09:55 AM.
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  3. #3
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    The only solution is (0,0,0) (over both the integers and the naturals).


    Sketch of proof:

    Since we're dealing with squares here, we can assume the integers are non-negative.

    From what Tarask did, we're left with showing that 2x^2+5y^2=z^2 has no integer non-negative solutions.

    From z^2\equiv 2x^2 \pmod{5} we conclude, by first checking what a square can be modulo 5,
    that 5\mid x and 5\mid z.

    Put this in the equation, check that 5\mid y and replace y with 5y'.

    Now we have the same equation as before, and thus we can conclude by Fermat's infinite descent that the only

    solution is (0,0,0).
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  4. #4
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    Quote Originally Posted by Unbeatable0 View Post
    The only solution is (0,0,0) (over both the integers and the naturals).


    Sketch of proof:

    Since we're dealing with squares here, we can assume the integers are non-negative.

    From what Tarask did, we're left with showing that 2x^2+5y^2=z^2 has no integer non-negative solutions.

    From z^2\equiv 2x^2 \pmod{5} we conclude, by first checking what a square can be modulo 5,
    that 5\mid x and 5\mid z.

    Put this in the equation, check that 5\mid y and replace y with 5y'.

    Now we have the same equation as before, and thus we can conclude by Fermat's infinite descent that the only

    solution is (0,0,0).
    I knew that we have to use FID , nice proof
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