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Fermat's equation solution in Z

  1. Feb 7, 2012 #1
    hi,

    is there integers a, b, c that satisfy the equation an + bn = cn for n>2
    (I don't mean positive integers)

    thx.
     
  2. jcsd
  3. Feb 7, 2012 #2

    mathman

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    I assume you mean an + bn = cn. Even if a, b, or c is negative, it won't make any difference. If n is even, it is obvious, while if n is odd you can replace the negative by its positive counterpart and rearrange terms so the equation is in its original form.
     
  4. Feb 7, 2012 #3
    There are many solutions where one of the numbers is 0: two examples are 64+04=64 and 53+(-5)3=03.

    There can be no solution with non-zero numbers, and this can be inferred from non-existance of positive solutions.
    For even n, this is trivial: since an=(-a)n, a solution with negatives is also a solution with positive integers.
    For odd n, (-a)n=-an, and you can always rearrange the terms to have a solution in positive numbers. This is a bit different for each assignment of signs to a, b, and c, but as an example, if a<0,b>0,c>0 and an+bn=cn, then cn+(-a)n=bn is a solution in positive numbers.
     
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