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Goodbye, Fermat’s theorem!

  1. Jan 2, 2007 #1
    I had lost: the target date is up! Here is farewell miniature

    It is common knowledge:
    If A^n+B^n=C^n, and ABC does not divided by prime n>2, then, according to Little Fermat’s theorem, in base n for number A, B, C there exist next equalities:

    1a°. A^n=(C-B)P, где C-B=a^n, P=p^n, A=ap;
    1b°. B^n=C^n-A^n=(C-A)Q, где C-A=b^n, Q=q^n, A=aq
    1c°. C^n=A^n+B^n=(A+B)R, где A+B=c^n, R=r^n, A=ar
    and the numbers
    2°. A+B-C=ap+bq-cr=u’n^k, where k>0 (corollary from Little Fermat’s theorem).
    3°. (A+B)-(C-B)-(C-A)=2u’n^k, where numbers (A+B), (C-B), (C-A) are composite.

    Then, after transformation of the last digit in the number c into n-1 (with help of multiplication of Fermat’s equality by corresponding number d^n), from 2°-3° the next equalities follow:
    4a°. a*x=(c*-b*)x,
    4b°. b*y=(c*-a*)y,
    4c°. c*z=(a*+b*)z
    where a*, b*, c* are the last digits in the numbers a, b, c,
    x, y, z are the last digits in the numbers p, q, r [and numbers (c^n-b^n)/(c-b), (c^n-a^n)/(c-a), (a^n+b^n)/(a+b)].
    On rewrite 3° as:
    5°. (a*+b*)z-(c*-b*)x-(c*-a*)y=a*z+b*z-c*x+b*x-c*y+a*y=2u’n^k, or
    a*(z+y)+b*(z+x)-c*(x+y)= 2u’n^k.
    Comparing 5° and 4°, we have:
    x=z+y, y=z+x, z=x+y, whence it follows that x=y=z=0. But according to Little Fermat’s theorem, x=y=z=1. Therefore, Fermat’equality has no positive solutions.

    If c* (or b*) is equal to zero, then also x=y=z=0.

    Happy New Year!
  2. jcsd
  3. Jan 2, 2007 #2

    matt grime

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    Science Advisor
    Homework Helper

    The first line, 1ao contains something completely unjustified.Why should A^n factor as (C-B)p^n, with C-B an n'th power too? What has FLT got to do with that? What is p? Nothing there requires at any point that n>2, by the look of it. (It would be nice if you were to highlight the points where you use the hypotheses of the theorem.)
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