Goodbye, Fermat’s theorem!

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!
 

matt grime

Science Advisor
Homework Helper
9,394
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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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