Fermat's last theorem is: contradiction on the parity

[Victor Sorok]

Well, my first objection so far is that you are assuming that a is a generator for some large q. How do you know that there is such a q? This cannot be proven using Dirischlet's theorem. =Uncool-

Существуют два доказательства теоремы о бесконечности множества чисел

 

There are two proofs of the theorem about infinity of the set of the numbers

[math]N(q)[/math].

++++++++++++++

 

Step back: the replacement of assertions 5-6 to the more general assertion:

 

5) Set [math]N(q)[/math] contains such digit [math]e2[/math] that

one half ([math]N'(q)[/math]) of set [math]N(q)[/math] is described by formula [math]e2^t_1[/math] ([math]t=1, 2... (q -1)/(2n)[/math]), another half ([math]N''(q)[/math]) of set [math]N(q)[/math] is described by formula [math](q-e^t_1)_1[/math] ([math]t=1, 2... (q -1)/(2n)[/math]).

Example of [math]N(q)[/math] for [math]n=5[/math]:

 

[math]N(31)[/math]: 1, 5, 6, 25, 26, 30;

[math]e=5[/math];

[math]N'(31)[/math]: 1 ([math]=5^0[/math]), 5 ([math]=5^1[/math]), 25 ([math]=5^2[/math]);

[math]N''(31)[/math]: 30 (31-1), 26 (31-5), 6 (31-25).

 

Two, of course, was more conveniently...

[Victor Sorok]

Only result after lately.

 

In all noted by me equalities [math]A+B=C[/math] on the last digits [math]A, B, C[/math] of the numbers [math]d^p, d^q, d^r[/math] the numbers [math]A[/math] and [math]C[/math] have common divisor.

If this fact is regular and it is easy to prove it, then simple proof of FLT exists.