Victor Sorok

Before shutting after itself door… From the identity (with a+b-c=0) (1°) (a+1)(b+1)(c+1)=abc+c^2+1 it follows that (2°) (a+1)(b+1)(c+1)-c^2=abc+1. In the Fermft’s equality the role of the numbers a, b, c play the numbers a^n, b^n, c^n. After the multiplication of Fermft’s equality by the sufficiently large number (3°) ones into 2° can be disregarded, and (4°) 0<<ab-c<<ab. And the equality 2° is converted into the inequality.

Proof of the FLT in the binary system of numeration for the prime n> 2. Designations: a_(i) – the i-th number from the end in the number a. a_(p, q) – the number, comprised of the digits of the number a from the rank p to the rank q. *** (01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 0 in positive integer numbers exists. (02°) While it is easy to show that A+B>C>A>B>U>0, where U=A+B-C. Let us lead the number U (in the binary system) to the form (03°) u =Ud=(2^r)(2^p-1) with the aid of the multiplication of the equality of 01° by the appropriate number d^n (which, as is known, exists), as a result what the equality 01° is converted into the equality (1°) a^n+b^n-c^n = 0 where, as is well known of the theory of the Fermat’s equality, (2°) a+b>c>a>b>u>0, c>a>c(n-1)/n; (3°) u=a+b-c=u_(q, r+1)(2^r); (4°) c-a=b-u=g>0; (5°) b/u<2; (6°) (c-a)/a<1/n. *** Let us break all ranks of the number u into three intervals: 1) [p, q+1], 2) [q, r+1], 3) [r, 1], where p – is the greatest rank of the number c, q – is the greatest rank of the significant part of the number u, r – is the greatest rank of the zero end of the number u. I.e. the number (7°) u = u_(p, q+1)(2^q)+u_(q, r+1)(2^r)+u_(r, 1) = /00…00//11…11//00…00/, and the equality a+b-c=u takes approximately this form: a= /1…11//11…10//x…/, + b= /0…01//00…01//y…/, - c= /1…10//00…01//z…/. ============== u=/0…00//11…11//00…00/. Let us carry out the digital analysis of this equality (principal moment of the proof). (8°) The number b_(p, q+1) can be equal only to 1 (which follows from 4° and 6°). (9°) If the digit a_(q+1)=1, then follows from the equality c_(p, q+2)=a_(p, q+2) that a>c, that contradicts to 2°. (10°) But if a_(q+1)=0, then c_(p, q+1)-a_(p, q+1)=1 (here 1 is a digit of the rank q+1). A difference b-u, as can be seen from a numerical example, is considerably less than 2^q. And in this case we have contradiction with 5°. But since the third it is not given, equality 1° does not have a solution. FLT is proven. *** P.S. However, the proof remains true and in the prime base n> 2. Thus, after small correction the previous proof is true.

Equality-double in FLT. (Passed moment of the rejected idea) Beginning is identical: Designations: a_(i) – the i-th digit from the end in the number a in the prime base n> 2. a_(p, r) – the number, comprised of the digits from p to r of the number a. 9 – designation of digit n-1, 8 – designation of digit n-2. Proof of FLT (01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists. (02°) Then, as it is easy to show, A+B>C>A>B>U>0, where U=A+B-C. Let us lead (in prime base n) the number u=a+b-c to the form: (03°) u =Ud= [n^(p-r)-1)](n^r) with the aid of the multiplication of the equality of 01° by the corresponding number d^n (which, as is known, in the prime base n exists), as a result of which the equality of 01° is converted into the equality (1°) a^n+b^n-c^n = 0, where a+b>c>a>b>u>0 and u=a+b-c. Let us break all digits of the number u into three intervals: 1) [p, q+1], 2) [q, r+1], 3) [r, 1], where p is greatest rank of the number c, q is the greatest rank of the significant part of the number u, r is greatest rank of the zero ending of the number u. I.e. the number (04°) u = u_(p, q+1)n^q+u_(q, r+1)n^r+u_(r, 1) = /00… 00//99… 99//00… 00/. Let us examine first the case a_(R, 1)+b_(R, 1)-c_(R, 1)=0 Is attained the following process: Let us replace all digits in the numbers a, b, c before (i.e. it is less) the range p by zero. We will obtain the numbers a*_(p, p), b*_(p, p), c*_(p, p) (here it is below for simplicity of formulas zero endings equal for all numbers we do not write! Either it is convertible all numbers into n-mal fractions with the integer parts of a_p, b_p, c_p) …and value of function F (a*, b*, C*), or F_ (p). Then let us enter in the numbers a*, b*, c* the digits of the previous (i.e. the smaller) range p -1 and let us calculate value F_ (p, p -1) with the values of the numbers a_(p, p -1), b_(p, p -1), c_(p, p -1). Then let us enter in the numbers a*, b*, c* the digits of range p -2 let us calculate value F_(p, p-2) with the values of the numbers a_(p, p-2), b_(p, p-2), c_(p, p-2). And so on - up to the restoration of all digits of the numbers a, b, c. *** Taking into account the formula of 4°, it is easy to see that the value (5°) F_(p, q+1)= a_(p, q+1)^n+b_(p, q+1)^n-c_(p, q+1)^n<0. (6°) In the interval from (p, q) to (p, r+1) function F(a*, b*, C*) it will be strictly increasing. (7°) A in the interval from (p, r+1) to (p, 1) function F (a*, b*, C*) it will be as a whole of that diminishing. *** Thus, function F (a*, b*, C*) in the closed interval [p, 1] in the first subinterval appears as a whole of that diminishing, in the second subinterval - strictly increasing, on the third - as a whole of that diminishing. But the hence it follows that extrapolated diagram of the function F (a*, b*, C*) from (a*_(p, p), b*_(p, p), c*_ (p, p)) to (a, b, c) has also THE FRACTIONAL solution (a' , b' , c'), moreover 0<a'<a,0<b'<b, 0<c'<c!!! There is above how to think over! Continuation follows. ++++++++++++++++ Interpretation and the conclusion: Either the numbers a, b, c are not given to the canonical form or (a^n)_(1) = -a_(1), (b^n)_(1) = -b_(1), (c^n)_(1) = -c_(1), that contradicts Little fermat's theorem. “The zero shot”: n=2 In this case (- a^n) _ (1) = a_ (1)=1, (- b^n) _ (1) = b_ (1)=1, (- c^n) _ (1) = c_ (1)=1, and there is no contradiction. Now it is possible to study the formulation of text. Simplified version of the proof Let us assume that the equation (1°) x^n+y^n-z^n = 0 has the entire solution (x, y, z) in the prime base n. Let us designate the last digits of the numbers x, y, z, through a, b, c and the number x, y, z, in which the numbers a, b, c are substituted with zero, through A, B, C, i.e., (2°) A=x-a, B=y-b, C=z-c. Proof of FLT Let us rewrite equality (1°) in the form: (3°) (A+a)^n+ (B+b)^n- (C+c)^n = 0, or (4°) (A^n+A') + (B^n +B')- (C^n +C') = 0, or (5°) (A^n+B^n-C^n) + (A' +B' - C') = 0, where: (6°) A^n+B^n-С^n>0 (<0), if a+b-c=0 (if a+b-c=1), (7°) the numbers A' , B' , C' finish respectively to the digits a, b, c and (8°) the numbers A' +B' - C' < 0 (or >0), if a+b-c=0 (if a+b-c=1). Consequently, or A'<0, or B'<0, or C'>0. But then: or in the number A^n +A' last digit is equal to …0-a=n-a or in the number B^n +B' last digit is equal to …0-b=n-b or in the number C^n +C' last digit is equal to …0-c=n-c. Therefore, в одном из чисел A^n, B^n, С^n нарушается малая теорема Ферма. So, допустив наличие решения уравнения 1°, мы получили противоречие. Therefore, in one of the numbers A^n, B^n, С^n is disrupted Little Fermat's theorem. So, after allowing the presence of the solution of the equation of 1°, we obtained contradiction. FINISH One additional reason: Let the numbers x^n, y^n, z^n finish to the digits a, b, c. Then: From x^n+y^n-z^n= (2x^n+2y^n-2z^n) /2= [(x^n+y^n) - (z^n-y^n) - (z^n-x^n)]/2 and (!!!) the equality x^n+y^n-z^n=0 it follows that during the multiplication of the expression of x^n+y^n-z^n by positive number 2 the last digits a, b, c in the numbers of x^n, y^n, z^n are opposite to last digits -a, -b, -c in the identically equal to them numbers (x^n+y^n), (z^n-y^n), (z^n-x^n). It is not strange whether?

Proof FLT. Final text. Designations: a_(i) – the i-th digit from the end in the number a in the prime base n>2. a_(p, r) – the number, comprised of the digits of the number a from the rank p to the rank r. 9 – designation of digit n-1. (01°) let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists. (02°) Then it is easy to show, A+B>C>A>B>U>0, where U=A+B-C. Let us lead the number U (in the prime base n) to the form (03°) u =Ud=(n^k)(n^p-1) with the aid of the multiplication of the equality 01° by the appropriate number d^n (which, as is known, exists), as a result of which the equality of 01° is converted into the equality (1°) a^n+b^n-c^n = 0, (2°) where a+b>c>a>b>u>0, (3°) u=a+b-c, where the number u has r digits (but the highest digit of the number c let us designate by letter p) and, let us note, that (4°) the number u is EVEN. Let us try to find the digital solution of equation (3°) lower the r-digit endings are isolated from the high-range digits by two skew lines. Let (5°) a*=…?v//x…, + b*=…?v//y…, - c*=…?w//z… ========== u = …00//9… It is easy to see that the sum of the numbers 1. x+y is as the minimum equal to 9 2. Consequently, x is as the minimum equal to 9/2. 3. Consequently, z is as the minimum equal to 9/2. 4. Consequently, x+y is as the minimum equal to 9+9/2… And so on to the eventual result: 5. x=y=z=9. 6. Consequently, v is as the minimum equal to 1. However, on this the calculation of numbers to end does not can, since in this case u_(r+1) is not equal to zero. In order to remove this contradiction, is necessary number w_ (r+1) to make equal to one. Of course other numbers, except these two, can enter into the discharges of higher than the r-th; however, when they satisfy the requirements: their sum is even. But then the sum of all numbers, which figure in the Fermat’s equation, will be ODD. Consequently, and the number of a^n+b^n-c^n will be ODD, which is impossible Thus, FLT is proven completely and indisputably. Alas, the idea of proof proved to be hopeless. Long time out.

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.

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...

I wrote incorrectly. ++++++++++++++++++ Position until today Designation 1: [math]a_1[/math] - the last digit of any number [math]a[/math]. Here is the enumeration of the assertions, which are necessary for the proof FLT: 1) for the given ones of the prime number [math]n>2[/math] and the odd number [math]d[/math] there is an infinite set of prime numbers [math]q[/math] of the form [math]q=2pn+1[/math], where odd [math]p[/math] and [math]d[/math] are relatively prime. For the proof FLT we will take 2) the number [math]d[/math], which is been the greatest odd divisor of the number [math](abc)^n[/math] from the Fermat’s equality, and 3) the number [math]q>c^n[/math] (i.e. the numbers [math]a^n, b^n, c^n[/math] are one-digit). Designation 2: [math]N(q)[/math] – the set of different (without the alliterations) last digits in the numbers [math]t^n[/math] ([math]t=1, 2... q-1[/math]) in the prime base [math]q[/math]. Then 4)The number of different elements in the set [math]N(q)[/math] is equal [math](q-1)/n[/math]. 5) Set [math]N(q)[/math] contains number [math]2[/math]. 6) One half of set [math]N(q)[/math] is described by formula [math]2^t_1[/math] ([math]t=1, 2... (q -1)/(2n)[/math]), another half of set [math]N(q)[/math] is described by formula [math](q-2^t_1)_1[/math] ([math]t=1, 2... (q -1)/(2n)[/math]). Two examples of [math]N(q)[/math] for [math]n=3[/math]: [math]N(31)[/math]: 1, 2, 4, 8, 15, 16, 23 (=31-8), 27 (=31-4), 29 (=31-2), 30 (=31-1); [math]N(43)[/math]: 1, 2, 4, 8, 11 (=43-32), 16, 21 (=64-43), 22 [=43-21=43-(64-43)], 27 (=43-16), 32, 35 (=43-8), 39 (=43-4), 41 (=43-2), 42 (=43-1). Thus, in the base [math]q[/math] Fermat’s equality takes the interesting form: 7) [math](2^u_1 + 2^v_1-2^w_1)_1=0[/math] (or even [math](1 + 2^v_1-2^w_1)_1=0[/math]), where [math]2^u_1=a^n, 2^v_1=b^n, 2^w=c^n[/math]. However, it can prove to be insufficiently for the completion of the proof of my knowledge. I propose to you to complete proof FLT. Time out

1. Yes. 2. Correction: "(4°) Since according to lemma 2 (since one-digit number [math]a[/math] and number [math]m[/math] mutually prime), all one-digit ends in [math]t-1[/math] numbers [math]a^t[/math] - consequently, and the numbers [math]a^{x+tm}[/math] - are different, there are such values [math ]x+tm=r[/math] and [math ]x+tm=s[/math] that the one-digit ends in the numbers [math]a^r[/math] and [math]a^s[/math] are equal to [math]b[/math] and [math]c[/math]". But "consequently, and the numbers [math]a^{x+tm}[/math]" is not true. 3. This is not true. Thanks

News Well and now - the most essential COMPUTER fact in a matter of elementary proof FLT: On the set [math]N[/math] of last digits in the numbers [math]q^n[/math], where [math]q[/math] is prime and [math]n [>2][/mat] is a divisor of the number [math]q-1[/math], the equality [math]A+B=C[/math] only if [math]A=1[/math] (or [math]B=1[/math]) or [math]A=B[/math] . (Obviously, if prime [math]q>c^n[/math], then Fermat’e equality is impossible on the set [math]N[/math].) There remains only to explain to us computer fact.

THE HYPOTHETICAL PROOF OF FERMAT'S LAST THEOREM, based on the following two lemma-theorems Lemma 1: For any odd number [math]a[/math] there is an infinite set of such prime numbers [math]q>a[/math], that the numbers [math]a[/math] and [math]q-1 [=m][/math] are mutually prime. Lemma 2: All last digits in the numbers [math]a^t[/math] in the base [math]q[/math], where [math]t=1, 2, … q-1[/math], are different, i.e., compose the complete set of positive digits in the base [math]q[/math]. [Now there is no time to recall the proofs of these lemmas - this can be put off to the future. But it is worthwhile to look comments of “shwedka” and “tolstopuz” on the page http://lib.mexmat.ru/forum/viewtopic.php?t=8525&start=150 (in Russian).] Essence of the obtained contradiction of the FLT: in the base [math]q[/math] the number [math]a^n+b^n -c^n=1[/math], but not [math]0[/math]. Thus, let us assume that (1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and [math]a[/math] or [math]c[/math] also odd. Proof VTF (2°) for the odd number [math]3na^n[/math] (or for [math]nc^n[/math]) let us take the prime number [math]q>3na^n>c^n[/math], which satisfies the condition of lemma 1. Let us compile linear diophantus equation with the mutually prime parameters [math]n[/math] and [math]m=q-1[/math]: (3°) [math]n(x+tm)-m(y+tn)=1[/math] with the general solution [math](x+tm; y+tn)[/math]. (4°) Since according to lemma 2 (since one-digit number [math]a[/math] and number [math]m[/math] mutually prime), all one-digit ends in [math]t-1[/math] numbers [math]a^t[/math] - consequently, and the numbers [math]a^x+tm[/math] - are different, there are such values [math]x+tm=r[/math] and [math]x+tm=s[/math] that the one-digit ends in the numbers [math]a^r[/math] and [math]a^s[/math] are equal to [math]b[/math] and [math]c[/math]. But according to Little Fermat's theorem, one-digit ends in the numbers [math]a^n[/math], [math]a^{rn}[/math] and [math]a^{sn}[/math] in the base [math]q[/math] are equal to [math]1[/math], and then the one-digit end of the number (5°) [math]a^n+a^{rn}-a^{sn}[/math], i.e., [math]a^n+b^n-c^n=1[/math], but not [math]a^n+b^n-c^n=0[/math]. FLT is proven.

You are right, but this does not refer to the proof: A = 28 and B = 29 have no common factor. ++++++++ Unfortunately, it was possible to prove the lemma only on the condition that the numbers [math]a_2[/math] and [math]b_2[/math] have one (to Maxim two) prime base, what it is clearly insufficient for the proof FLT. Possibly, something can give the use of the fact that each prime cofactor of the numbers [math]V, a_2[/math] and [math]b_2[/math] take the form [math]en+1[/math]…

In the Fermat’s equality all numbers a, b, c more then n^3.

1. Yes, every pair of numbers is relatively prime. 2. Please repeat the error which you found. In the march after the lemma! I until did not find the simple proof of lemma. But here one of the ideas of its proof under the conditions VTF. Is created impression, that remainder from the division of the number [math]a_2[/math] into [math]V[/math] less [math]a_1[/math], and remainder from the division of the number [math]b_2[/math] into [math]V[/math] less [math]b_1[/math]. IF THIS IS SO, then the truth of lemma is obvious.

Note: In new texts [math]A=a, B=b, C=c, a'=a_1, a''=a_2[/math], etc. It is not true if, for example, c is divided by n: Then the numbers [math]a+b[/math] and [math]R[/math] are divided by [math]n[/math] and the formulas 2° for [math]a+b[/math] and [math]R[/math] is not true. From 2°: [math] a_1^n =c-b, P=a_2^n[/math], [math] b_1^n =c-a, Q=b_2^n[/math], [math] c_1^n =a+b, R=c_2^n[/math]: [math] a_1^n +b_1^n=(c-b)+(c-a)=2c-(a+b)=2c_1c_2-c_1^n[/math], etc.

==================================== Deviation in the proof after the formula 7°. ==================================== I remind basic properties of the Fermat’s equality (1°) [math]a^n+b^n = c^n[/math], where prime [math]n > 2[/math] and [math]a, b, c[/math] have no common factors. Let for the time being the number [math]abc[/math] is not divided by [math]n[/math]. Then: (2°) [math]c^n-b^n=(c-b)P[/math], where [math]c-b=a_1^n, P=a_2^n[/math], [math]c^n-a^n=(c-a)Q[/math], where [math]c-a=b_1^n, Q=b_2^n[/math], [math]a^n+b^n=(a+b)R[/math], where [math]a+b=c_1^n, R=c_2^n[/math], [this follows from the simple lemma: if the numbers [math]A[/math] and [math]B[/math] have no common factors and the number [math]A+B[/math] is not divided by [math]n[/math], then the numbers [math]A+B[/math] and [math]R = \frac{A^n+B^n}{A+B}[/math] have no common factors (since [math]R=(A+B)^2T+nAB)[/math]]; (3°) the numbers [math]a_1, a_2, b_1, b_2, c_1, c_2[/math] have no common factors, the numbers [math]P, Q, R[/math] and [math]a_2, b_2, c_2[/math] нечетны; (4°) [math]a=a_1a_2, b=b_1b_2, b=c_1c_2[/math]. (5°) From analysis of difference of the polynomials [math]P[/math] and [math]Q[/math] in its standard (school) form one can see, that the number [math]P-Q[/math] is divided by [math]a-b[/math]. (6°) Important equalities: [math]c_1^n - b_1^n = 2a_1a_2 - a_1^n= 2a - (c-b)[/math], [math]c_1^n - a_1^n = 2b_1b_2 - b_1^n = 2b - (c-a)[/math], [math]a_1^n + b_1^n = 2c_1c_2 - c_1^ n =2c - (a+b)[/math], from which we find the base formula of the number [math]g=a_2 - b_2[/math]: (7°) [math]g= a_2 - b_2 = \frac{(a+b)(b_1-a_1) + (a_1+b_1)( a_1^n-b_1^n)}{2a_1b_1}[/math], from which one can see, that the number [math]g=a_2-b_2[/math] is divided by [math]a_1-b_1[/math]. ========================================== Next we write down the numbers [math]P-Q[/math] and [math]a-b[/math] in the form: (8°) [math]P-Q= a_2^n-b_2^n=(a_2-b_2)W[/math] and (9°) [math]a-b= a_1^n-b_1^n=(a_1-b_1)V[/math]. Now, comparing its efficient in (5°) and (7°), one can make interesting conclusion: the number [math] a_2^n-b_2^n=(a_2-b_2)W[/math] is divided by [math]a_1^n-b_1^n(a_1-b_1)V [/math], but the number [math]a_2-b_2[/math] is divided only by [math]a_1-b_1[/math]. Consequently, the number [math]W[/math] is divided by [math]V[/math]. BUT: if the numbers [math]a_1, b_1, a_2, b_2[/math] have no common factors (this follows from 2°) the numbers [math]W[/math] and [math]V[/math] have NO common factors. (It is not inconceivable that this assertion constitute well-known theorem from the theory of discontinuous numbers. And if this lemma-theorem is known in mathematics, then elementary proff of FLT is on hand.) Numerical example to lemma-theorem. Let [math]n=3, a=1, b=4, c=2, d=3, a+b=5, c+d=5[/math], [math]a+b[/math] is divided by [math]c+d[/math], but [math]\frac{a^n+b^n}{a+b}=13[/math] is NOT divided by [math]\frac{c^n+c^n}{c+c}=7[/math]. Victor Sorokine. Mezos (France) ======================== And since it is evident from the previous post, difficulty of FLT is a difficulty of final lemma-theorem. Possibly, someone will give reference to its publication. Possibly, someone will be able to give its (judging by the analogous theorems) simple proof. Well and no - it is necessary to search for proof. The first (previous) numerical example illustrates the essence of theorem. But here is more strongly example for the [math]a, b, c, d[/math], which have no common factors. Let [math]c+d[/math] is divided by [math]a^n+b^n[/math]. Then [math]c+d[/math] is divided by [math]a+b[/math]. But [math]\frac{c^n+d^n}{c+d}[/math] and [math]c+d[/math] have no common factors (cf. notice to 2°). Therefore, [math]\frac{c^n+d^n}{c+d}[/math] is NOT divided by [math]\frac{a^n+b^n}{a+b}[/math], i.e. [math]W[/math] is NOT divided by [math]V[/math]. For proof lemma-theorem in genera case maybe one can use the fact that - in conditions of FLT – the numbers [math]a_2[/math] andи [math]b_2[/math] little differ one from other. There are many other considerations. Lemma-theorem (consequently and VTF) can be brought to the different simple theorems. Here one of them: The numbers [math]\frac{a^n-b^n}{a-b}[/math] and [math]\frac{(a+1)^n-(b+1)^n}{a-b}[/math], where [math]a-b[/math] is not is divided by [math]n[/math] (and numbers [math]a[/math] and [math]b[/math] have no common factors), have no common factors. What will say in regard to this the specialists in the theory of the numbers? And... what will say schoolboys about actually five-line proof of FLT (or its reducing to lemma-theorem) in 8°-9°?

I had proved that If a solution of the Fermat’s equality in whole numbers exists, then the number [math]g = a_2-b_2[/math] is divided by [math]4n(a_1 - b_1)[/math], i.e. the number [math]h= \frac{a_2-b_2}{4n(b_1-a_1)}[/math] is whole. On the other hand, calculations on computer for real number [math]a, b, c[/math], which answer to the Fermat’s equality, show that [math]0 < h << 1[/math]. Question: is it possible to show this fact? V.S.

Since a+b= 2^kn * x and c= 2^k * x BUT! 1. Alas, and this proof proved to be incorrect in PRINCIPLE. 2. However, if elementary proof FLT exists, then it is located somewhere entirely next. I analyze errors and new properties of Fermat’s equality. Sometimes I will present the logic of this analysis, and the thus far basic conclusion: contradiction must be revealed only with the aid of THE INEQUALITIES. 3. Long ago I advanced this hypothesis: since the number [math]P-Q[/math] is divided by [math]A-B[/math] and, moreover, by common divisor of the numbers [math]c[/math] and [math]a+b[/math], then in number [math]P-Q= a''^n -b''^n[/math] precisely cofactor [math]a''- b"[/math] is divided by [math] (A-B)c'[/math]. But prove this for me everything could not. 4. On 2 August I proposed proof with the aid of the number [math]a''-b'' = \frac{(b'-a')(A+B) + (a'+b')(a'^n-b'^n)}{2a'b'}[/math] (cf. formula (7a°) #1 on 08-11-2007, 09:39 PM). And here only today I noted that number [math]a''- b''[/math] is divided also by [math] (A-B)c'[/math]. 5. Now it is possible to show that the number [math]\frac{ a''-b''}{(A-B)c'}[/math] is IMPROPER FRACTION, that contradicts p.3. (Thus, to me it is obvious that [math]a''-b''< (\frac{P-Q}{a''-b''})^{\frac{1}{n}}< c'[/math] etc.) Thus, I gradually approach the simple calculations. Splendid confirmation of the hypothesis! Calculation of the difficultest case: [math]n=3[/math], the numbers [math]A and B[/math] are not divided by [math]n[/math]. If [math]A=100[/math] and [math]B=200[/math], then [math]a''-b''=7,9834[/math] and[math](a'-b')c' = 18,476[/math]. Therefore the fraction [math]\frac{a''-b''}{(a'-b')c'}[/math] is NOT whole! ++++++++++++++++ Calculations: [math]a=100[/math], [math]b=200[/math], [math]c=208,008[/math]; [math]a^3=1000000[/math], [math]b^3=8000000[/math], [math]c^3=9000000[/math]; [math]c-b=8,008[/math], [math]c-a=108,008[/math], [math]a+b=300[/math]; [math]a_1=2,001[/math], [math]b_1=4,762[/math], [math]c_1=6,694[/math]; [math]a_2=49,983[/math], [math]b_2=41,999[/math], [math]c_2=31,074[/math]; [math]b_2-c_2=10,925[/math], [math]a_2-c_2=18,909[/math], [math]a_2-b_2=7,983[/math]; [math]c_1+b_1=11,456[/math], [math]c_1+a_1=8,695[/math], [math]b_1-a_1=2,761[/math]; [math] (c_1+b_1)a_1[/math]: [math](c_1+a_1)b_1[/math]: [math] (b_1-a_1)c_1[/math]: [math] 22,923[/math]; [math] 41,406[/math]; [math]18,482[/math]. Another confirmation of the hypothesis: If [math]n=3[/math], [math]a=180[/math] и [math]b=200[/math], then we have: [math]a_2-b_2=11,79[/math], but [math] (b_1-a_1)4n=14,844[/math] (i.e. 11,79<14,844), [math]c_2+b_2=37,287[/math], but [math] (c_1+a_1)4n=134,388[/math] (i.e. 37,287<134,388), [math]ca_2+a_2=25,497[/math], but [math] (c_1+a_1)4n=149,232[/math] (i.e. 25,497<149,232). +++ It is obvious, when [math]a[/math] approach to [math]b[/math], then [math]f = (b_1-a_1)4n[/math] approach to [math]g = a_2-b_2[/math]. Apropos, from formula 7° it follows that [math]g = a_2 - b_2[/math] is divided by [math]4n(a_1 - b_1)[/math], but – as calculations show in concrete numbers - (8°) [math]g = a_2 - b_2 < 4n(a_1 - b_1)[/math]. It is the CONTRADICTION of the Fermat’s equality.

Elucidating trifles 1. Formulation of the contradiction: Sum and difference in the odd numbers cannot simultaneously be divided into 4. 2. Details about the [math]i[/math]-th substitution: (3-i°) [math]a_{i-1}=c_i-b_i, b_{i-1}=c_i-a_i, c_{i- 1}=a_i+b_i[/math], where (4-i°) [math]a_i+b_i-c_i=2^{k-i-1}x[/math]. It is easy to see that (5-i°) [math]a_i+b_i=2^{k-i}x[/math], [math]c_i=2^{k-i-1}x[/math], and the numbers [math]a_i, b_i, c_i[/math] are whole, but the numbers [math]a_i[/math] and [math]b_i[/math] are odd. Thus, after the latter ([math]k[/math]-th) substitution all numbers [math]a_k, b_k, c_k[/math] become odd. 3. The case with even [math]a[/math] proves in perfect analogy, but with the only difference that equal in the absolute value items [math]y[/math] in the numbers [math]c_k[/math] and [math]b_k[/math] have the opposite signs. And now after the substitution of the values [math]c[/math] and [math]b[/math], expressed through the numbers [math]c_k[/math] and [math]b_k[/math], we obtain the contradiction: both numbers [math]c^n-b^n[/math] and [math] c^n+b^n [/math] are divided by [math]4[/math], that - if [math]c^n[/math] and [math]b^n[/math] are odd – it is impossible. Corr: (5-i°) [math]a_i+b_i=2^{k-i+1}x[/math], [math]c_i=2^{k-i}x[/math], and the numbers [math]a_i, b_i, c_i[/math] are whole, but the numbers [math]a_i[/math] and [math]b_i[/math] are odd.

Turning (after MINI-proof) in the of the FLT. Complete text from the very beginning... Thus, let (1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and (for the beginning) [math]a[/math] and [math]b[/math] are odd and [math]c=2^kx[/math], where here and here and everywhere further the number [math]x[/math] is odd, value of which DOES NOT HAVE an importance. It is easy to see that (2°) [math]a+b-c=2^kx[/math]. Let us show that equality 1° in the wholes numbers is insoluble. Let us make the first substitution: (3-1°) [math]a=c_1-b_1, b=c_1-a_1, c=a_1+b_1[/math], where (4-1°) [math]a_1+b_1-c_1=2^{k-1}x[/math]. It is easy to see that (5-1°) [math]a_1+b_1=2^kx[/math] and the numbers [math]a_1, b_1, c_1[/math] are whole, but the number [math]a_1[/math] and [math]b_1[/math] are odd. Let us make the second substitution: (3-2°) [math]a_1=c_2-b_2, b_1=c_2-a_2, c_1=a_2+b_2[/math], where (4-2°) [math]a_2+b_2-c_2=2^{k-2}x[/math]. It is easy to see that (5-2°) [math]a_2+b_2=2^{k-1}x[/math] and the numbers[math]a_2, b_2, c_2[/math] are whole, but the numbers [math]a_2[/math] and [math]b_2[/math] are odd. ………………………………………. And so on to [math]k[/math]-th substitution: (3-k°) [math]a_{k- 1}=c_k-b_k, b_{k-1}=c_k-a_k, c_{k- 1}1=a_k+b_k[/math], where (4-k°) [math]a_k+b_k-c_k=2^kx[/math]. It is easy to see that (5-k°) [math]a_k+b_k=2^1x[/math] and the numbers [math]a_k, b_k, c_k[/math] are whole, but the numbers [math]a_k[/math] and [math]b_k[/math] are odd. But then from 4-k° we have: (6°) [math]a_k-b_k=2^2y[/math], where [math]y[/math] is whole number, value of which DOES NOT HAVE an importance. Now let us express the numbers [math]a, b, c[/math] через [math]a_k, b_k, c_k [/math]. At that equal items in the numbers [math]a[/math] and [math]b[/math] we will reject, since to us is important only the difference of the numbers [math]a[/math] and [math]b[/math]. As a result, obviously, we obtain the equalities: (7°) [math]a=b_k+x[/math] and [math]b=a_k+x[/math] (or [math]a=a_k+x[/math] and [math]b=b_k+x[/math]), where [math]x[/math] – rejected the number. And now, after calculating the number [math]a^n-b^n[/math], we find that it IS MULTIPLE FOUR. But, as can be seen from 1°, and the number [math]a^n-b^n[/math] is also MULTIPLE FOUR, that by odd numbers [math]a^n[/math] and [math]b^n[/math] is impossible. FLT is proved.

Mini-proof of the case [MATH]a'+b'[/MATH] is divided exact by two: Then numbers [MATH]a'- b'[/MATH], therefore, and [MATH]a'^n-b'^n[/MATH] or [MATH]A-B[/MATH], therefore and [MATH]A^n-B^n[/MATH], are divided by or [MATH]4[/MATH]. But sum and difference of odd numbers [MATH]A^n[/MATH] and [MATH]B^n[/MATH] cannot simultaneously be divided into [MATH]4[/MATH]. And contradiction is present.

Complet text of the proof All wholes numbers in the next are given in the prime base [math]n[/math]. Here are the known positions, repeatedly discussed on the mathematical forums. Thus, let (1°) [math]A^n+B^n = C^n[/math], where prime [math]n > 2[/math] and [math]A, B, C[/math] have no common factors. Let for the moment the number [math]ABC[/math] is not divided by [math]n[/math]. Then: (2°) [math]C^n-B^n=(C-B)P[/math], where [math]C-B=a'^n, P=a''^n[/math], [math]C^n-A^n=(C-A)Q[/math], where [math]C-A=b'^n, Q=b''^n[/math], [math]A^n+B^n=(A+B)R[/math], where [math]A+B=c'^n, R=c''^n[/math], [this follows from the simple lemma: if the numbers [math]A[/math] and [math]B[/math] have no common factors and the number [math]A+B[/math] is not divided by [math]n[/math], then the numbers [math]A+B[/math] and [math]R = \frac{A^n+B^n}{A+B}[/math] have no common factors (since [math]R=(A+B)^2T+nAB)[/math]]; (3°) the numbers [math]a', a'', b', b'', c', c''[/math] have no common factors; (4°) [math]A=a'a'', B=b'b'', C=c'c''[/math]. (5°) Important equalities: [math]c'^n - b'^n = 2a'a'' - (C-B)[/math], [math]c'^n - a'^n = 2b'b'' - (C-A)[/math], [math]a'^n + b'^n = 2c'c'' - (A+B)[/math], or (5°a) [math](A+B) - b'^n = 2a'a'' - a'^n [/math], [math](A+B)- a'^n = 2b'b'' - b'^n [/math], [math]a'^n + b'^n = 2c'c'' - (A+B)[/math], from here we find the numbers [math]a''[/math] and [math]b''[/math]: (6°) [math]a'' = \frac{(A+B) - b'^n + a'^n}{2a'}[/math], [math]b'' = \frac{(A+B) - a'^n + b'^n}{2b'}[/math]. At last, we find two important formulas, лежащие in the base of the proof of the FLT: (7a°) [math]a'' - b'' = \frac{(b'-a')(A+B) + (a'+b')(a'^n-b'^n)}{2a'b'}[/math], (7b°) [math]a'' + b'' = \frac{(b'+a')(A+B) - (a'-b')(a'^n-b'^n)}{2a'b'}[/math]. Elementary proof of the Fermat's last theorem Let for distinctness the numbers [math]A[/math] and [math]B[/math] is not divided by [math]n[/math]. It is easy to see that the formulas 7° remain correct and if the number [math]C[/math] is not divided by [math]n[/math]. Let us examine the last two-three significant digits in the numbers [math]a'', b'', P, Q[/math] in binary system. Here are still the possible cases: a) if [math]a'=[/math]…[math]01[/math], [/math]b'=[/math]…[math]01[/math] [or [math]a'=[/math]…[math]11, b'=[/math]…[math]11[/math]], then [math]a''+ b''=[/math]…[math]00[/math], but [math]P+Q=[/math]…[math]10[/math]; b) if the number [math]a'+b'[/math] has [math]k[/math] zeros on the end, then the number [math]a''-b''[/math] has [math]k[/math] zeros on the end, but the number [math]P-Q[/math] has [math]k+1[/math] zeros on the end [since the number [math]P-Q[/math] is divided by [math]a-b[/math] and, moreover, by common divisor of the numbers [math]c[/math] and [math]a+b[/math]]; c) if [math]a'=[/math]…[math]0[/math] [or [math]b'=[/math]…[math]0[/math]], then both numbers [math]a''- b''[/math] and [math]a''+ b''[/math] are divided by [math]4[/math], that is impossible since the numbers [math]a''[/math] and [math]b''[/math] have no common factors. If the number [math]A[/math] is divided by [math]n[/math], then they are taken the numbers [math]c'', b'', R, Q[/math]. Their analysis leads to the analogous contradictions: (8a°) [math]c'' - b'' = \frac{(c'-a')(C+B) - (c'+b')(C-B)}{2a'b'}[/math], (8b°) [math]c'' + b'' = \frac{(c'+a')(C+B) - (c'-b')( C-B)}{2a'b'}[/math]. Takeing into account school formulas for polynomials [math]R[/math] and [math]Q[/math], , it is easy to see that the number [math]R - Q[/math] is divided by even number [math]C+B[/math] and, furthermore, to the common divisor even numbers [math]A[/math] and [math]C-B[/math]. Consequently, number [math]C+B[/math] is divided by 2 only to the first power. Let us examine the last two-three significant digits in the numbers [math] c'', b'', R, Q [/math] in binary system. Here are still the possible cases: a) if [math]c'=[/math]…[math]01, b'=[/math]…[math]01[/math] [or [math]c'=[/math]…[math]11, b'=[/math]…[math]11[/math]], then [math]c''+ b''=[/math]…[math]00[/math], but [math]R+Q=[/math]…[math]10[/math]; b) if the number [math]c'+b'[/math] has [math]k[/math] zeros on the end, then the number [math]c''-b''[/math] has [math]k[/math] zeros on the end, but the number [math]R-Q[/math] has [math]k+1[/math] zeros on the end [since the number [math]R-Q[/math] ] is divided by [math]C-B[/math] and, moreover, by common divisor of the numbers [math]A[/math] and [math]C+B[/math]]; c) if [math]c'=[/math]…[math]0[/math] [or [math]b'=[/math]…[math]0[/math]], then both numbers [math]c''- b''[/math] and [math]c''+ b''[/math] are divided by [math]4[/math], that is impossible since the numbers [math]c''[/math] and [math]b''[/math] have no common factors. (It is necessary to keep in mind that the number [math]R[/math] [is analogous and [math]P, Q[/math]] is even only if both number [math]A[/math] and [math]B[/math] are even.)Thus, FLT is completely proven. Victor Sorokine Mezos, 2 August 2007

Lemma is not true. New version of the lemma: If in the simple base q the mutually-simple numbers A^p, B^p, C^p (where C=c^n, B=b^n, A=a^n, prime n>2 and p=/q-1) finish by digit 1, then either A+B and A-B or C+B and C-B, or C+A and C-A have common divisor not equal to 1. Today I have not its proof. Time out

+++ Elementary proof of Fermat's last theorem According to known theorem from the theory of integers, for any prime [MATH]n[/MATH] there exist infinite great number of the prime numbers [MATH]d[/MATH] of form [MATH]d=pn+1[/MATH], where [MATH]p[/MATH] is integer. (This theorem is the simple consequence from the fact that, if numbers [MATH]a[/MATH] and [MATH]b[/MATH] are mutually-prime and [MATH]n[/MATH] prime, then each divisor of the number [MATH] (a^n+b^n)/(a+b)[/MATH] takes form [MATH]pn+1[/MATH]). Proof FLT: Let us take simple [MATH]d>c^{4n}[/MATH]. Then, according to Fermat's little theorem, the numbers [MATH]C^p-B^p, C^p-A^p, A^p-B^p[/MATH], where [MATH]C=c^n, B=b^n, A=a^n[/MATH], are divided by [MATH]d[/MATH]. And since [MATH]A+B=C[/MATH], then, according to Lemma, number [MATH]3C^2[/MATH] is divided by [MATH]d>3C^2[/MATH]. Consequently, the solution of Fermat’s equation is not integer.

Lemma. If integers [MATH]a, b, c[/MATH] have only one common divisor [MATH]1[/MATH], [MATH]a+b=c[/MATH], number [MATH](c^n-a^n)[/MATH] and [MATH](c^n-b^n) [/MATH] have common prime divisor [MATH]d>3c^2[/MATH], in base d numbers [MATH]a^n, b^n, c^n[/MATH] finish by digit [MATH]1[/MATH], then with [MATH]n>1[/MATH] either number [MATH](a^n-b^n)/(a+b) [/MATH] is not divided by [MATH]d[/MATH], or number [MATH]3c^2[/MATH] is divided by [MATH]d[/MATH]d. Proof. Let us assume that number [MATH](a^n-b^n)/(a+b)[/MATH] is divided by d. Then number [MATH]S=(c^n-b^n)/(c-b)+(c^n-a^n)/(c-a)+(a^n-b^n)/(a+b)[/MATH], or [MATH] (c^n-b^n)/a+(c^n-a^n)/b+(a^n-b^n)/c = [c^(n+2)+b^(n+2)+(a+b)c^(n+1]/abc=[/MATH] [MATH]=[c^2+b^2+a(c+b)]/abc[/MATH], is divided by [MATH]d[/MATH]. But since number [MATH]a^n, b^n, c^n[/MATH] finish by number [MATH]1[/MATH], then the number [MATH]s = c^2+b^2+a(c+b)[/MATH], where [MATH]0<s<d[/MATH], is single-digit number and, therefore, number [MATH]S[/MATH] is NOT DIVIDED by [MATH]d[/MATH].

Here is the Lemma, which easily proves and which is sufficient for the brief proof of the FLT: Lemma. If integers a, b, c have only one common divisor 1, a+b=c, number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided into d ". to be continued