Victor Sorokine

It seems that Pierre Fermat was rights…

Interesting fact: in the Fermat’s equality

U=(a+b)-(c-b)-(c-a)<2u=2(a+b-c)!

Let for integers a, b, c and odd n>2 there is a equality

(1°) a^n+b^n-c^n, where the number

(2°) u=a+b-c>0.

Then in the equivalent equality

(3°) (a+b)R-(c-b)P-(c-a)Q=0, where the numbers P, Q, R are known polynomials of the decomposition of the sum of two degrees,

the number

(4°) U=(a+b)-(c-b)-(c-a)=2u.

However, this equality is not carried out.

Are actual, with equal P, Q, R - for example, with

(5°) P=Q=R=a^(n-1) –

the equality of 4° is correct.

However, in comparison with the equality of 5° in the equality of 4° coefficient R attached to positive item (a+b) IS LESS a^(n-1) and coefficients P and Q attached to negative items - (c-b) and (c-a) ARE MORE a^(n-1).

As a result the number

(6°) U=(a+b)-(c-b)-(c-a)<2u=a+b-c.

And we have a contradiction since (a+b)-(c-b)-(c-a)=2(a+b-c).

It seems that Pierre Fermat was rights…

Let us assume that for the integers a, b, c there is a equality

(1°) a^n+b^n=c^n, where simple n> 2.

Let us examine the equality of 1° in the prime base

(2°) q=pn+1>3c^ {2n} (it is known that the set of such numbers q is infinite).

It is easy to see that the number

(3°) D=c^{qn} - a^{qn} - b^{qn} is divided by q. But

(4°) D=c^{npn+n} - a^{npn+n} - b^{npn+n}, where the numbers

(5°) c^{np}, a^{np}, b^{np} finish to digit 1 and, therefore, the sum of the last digits of the number D into 4° will be equal to q:

(6°) c^{2n} - a^{2n} - b^{2n} =q, that contradicts 2°.

(Sep 13, 2008)

In 1990 I proved FLT for the case when c is even.

Today I found proof, also, for even a /or b/. Here is it.

Designations:

p - an integer,

q - an odd number,

X - set of odd numbers of the type of x=2q+1,

Y - set of odd numbers of the type of y=4p+1.

Let us name two odd numbers x' and x'' uniform.

Let us name two odd numbers x and y diverse (different-type).

The following assertions are obvious:

(1°) if odd numbers a and b are of the same type, then the numbers a+2q and b are diverse.

(2°) the numbers of 2p+q and 2p- q are diverse.

Proof of FLT for the odd of n> 2.

Let us assume

(3°) a^n+b^n=c^n, or

(4°) (c-b)P+(c-a)Q=(a+b)R,

where two of the numbers a, b, c and of the numbers c-b, c-a, a+b are odd and

(5°) c>a>b>0.

Case 1. Number c is even.

Let us select the pair of the numbers of a+b=e (even) and of c-a=d (odd)

and let us examine the pair of numbers of d+e and d-e, which are been, obviously, DIVERSE (DIFFERENT-TYPE) (see 2°).

From the other side, THE SAME numbers of d+e (=c+b) and d-e (=c-b-2a) are UNIFORM, since the numbers of c+b and c-b are diverse, the numbers c-b and c-2a are also diverse, therefore, the number c+b and c-b-2a are uniform.

And we have a contradiction: the numbers in two identical pairs of numbers are different-type and at the same time uniform.

Case 2. Number a /or b/ even.

After the simplest substitution –

a=2^k-a*, where 2^k>c+a, –

the proof of this case for the numbers c, a*, b is completely analogous to previous proof.

(It is certain, - “this it cannot be according to the basic postulate of official science, because this it cannot be ever!” - my proof is not correct.)

Aug 26, 2008