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Math Help - Showing something is divisible

  1. #1
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    Showing something is divisible

    Hey there I need some help with this problem.

    If x^3 + y^3 = z^3, show that one of x,y,z is divisible by 7

    Just wondering if someone can show me how to do this problem, so I can understand how to do it thanks.
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  2. #2
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    Re: Showing something is divisible

    Consider this equality modulo 7.
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  3. #3
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    Re: Showing something is divisible

    The question assumes that x,y and z are integers. So, by Fermat's Last Theorum(FLM) there are no solutions to the equation, ie no solution divisible by 7.



    "FLM:There exists no solutions to the equation x^n +y ^n = z^n for n>2, x,y,z are integers and xyz≠0

    It is a strange question, or have I missed something?
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  4. #4
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    Re: Showing something is divisible

    note that 03 + 13 = 13 is a solution, and that 0 is divisible by 7.

    another solution is (-a)3 + a3 = 03 (again, 0 is divisible by 7).

    this is not a special case of FLT.

    emakarov's suggestion is pertinent, here, let's look at the cubes (mod 7):

    03 = 0 (mod 7)
    13 = 1 (mod 7)
    23 = 1 (mod 7)
    33 = 6 (mod 7)
    43 = 1 (mod 7)
    53 = 6 (mod 7)
    63 = 6 (mod 7).

    since x3 + y3 = z3, z3 = 0,1, or 6 (mod 7).

    since if z3 = 0 (mod 7) there is nothing to prove, assume that z3 = 1 or 6 (mod 7). then we have just two cases:

    z3 = 1 (mod 7).

    checking the possibilities (mod 7) we have:

    0+0 = 0
    0+1 = 1 <--this works, so x3 = 0 (mod 7), thus x = 0 (mod 7) (since the only thing (mod 7) when cubed that gives 0, is 0).
    0+6 = 6
    1+1 = 2
    1+6 = 0
    6+6 = 5

    (this is all we need to consider since addition mod 7 is commutative).

    for z3 = 6 there is also just one possibility:

    0+6 = 6

    here, again x3 = 0 (mod 7), so x = 0 (mod 7).
    Thanks from gfbrd
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  5. #5
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    Re: Showing something is divisible

    I see now thanks a lot for your help
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