Victor Sorokine

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