Victor Sorokine

FLT. Unproven case II (h=a+b-c=0) with integer d>1: [math]A^n+B^n=C^n[/math], or taking into account 2°-5° [math] (dU+a)^n+(U+b)^n-(dU+c)^n=0[/math], where integers 0<(a, b, c)<0.5U. Very interesting task. +++ First approach: Among the general solution [math](x+tmy, y+tax)[/math] of the linear diophantine equation [math]a(x+tmy)-m(y+tax)=1[/math] to find such a particular solution that, for example, [math]b(x+tmy)-mz=2[/math]. And now after multiplication Fermat’s equalitie by [math](x+tmy)^n[/math] the numbers A, B, C in the base m finish to 1, 2, 3 and the number [math]a^n+B^n-C^n[/math] finishe by [math]1^n+2^n-3^n[/math], but NOT by ZERO. Merged post follows: Consecutive posts mergedI have a Question: For relatively prime a and m the general solution of the linear diophantine equation ax-my=1 there will be, as is known, such: (x+tm, y+ta). Question: Are there such t and z, that b(x+tm)-mz=1, where a, b, m are relatively prime? Thanks

Really, -0.5=<a=<+0.5 -0.5=<b=<+0.5 -0.5=<c=<+0.5 +++++++++++ -1.5=<h=<+1.5, where h is WHOLE. In the interval [-1.5, 1.5] are only THREE integers.

Corrections: 2°) [math]U<B<1.5U[/math]. Actually, [math]from [b^n=] BB^(n-1)=(C-A)Q[/math] it follows: [math]BB^(n-1)>[/math] [math]>(C-A)[nB^(n-1)][/math] (see formula for Q). From here [math]B>3(C-A), B>3(B-U), 3U>2B, B<1.5U[/math]. ..... I. If h=1 and g=0, then [math](1+t+0,5)^n+(1+0.5)^n-(2+t+0)^n<max(H)=max(A^n+B^n-C^n)=[/math] [math]=(1+0.5)^n+(1+0.5)^n-(2+0)^n<-1.24<0[/math] with d=1, e=1, f=2; a=0.5, b=0,5, c=0 and any t>0. [With [math]n=2 max(H)=(1+0.5)^2+(1+0.5)^2-(2+0)^2=0.5>0[/math]!!! And equality H=0 is possible.] Merged post follows: Consecutive posts mergedProof of the Fermat’s Last Theorem (finish text) Let us assume that for integers A, B, C (C>A>B>0) and n>2 1°) [math] (H=) A^n+B^n-C^n=0[/math], where [math]A+B-C=U>0[/math] [is obvious, C>A>B>U] and 2°) [math]U<B<1.5U[/math]. Actually, from [math] [b^n=] BB^{n-1}=(C-A)Q[/math] it follows: [math]BB^{n-1}> >(C-A)[nB^{n-1}][/math] (see formula for Q). From here [math]B>3(C-A), B>3(B-U), 3U>2B, B<1.5U[/math]. The numbers A, B, C can be written down on the module U in two forms: with deficiency and with the surplus. For example, A=27 with the module u=12 can be written down thus: A=2U+3=3U-9. Of two versions of the numbers A, B, C let us leave those, in which the remainder (3 or 9) will be LESS: 2a°) A=dU+a, B=eU+b, C=fU+c, where 0<(a, b, c)<0.5U. It is easy to show (see the Appendix) that in the equality of 1° the numbers a, b, c are equal neither 0 nor U/2. From the equality U=A+B-C=(dU+a)+(eU+b)-(fU+c), or 3°) [math] (d+e+f)U+(a+b-c)=U[/math] follows that the number h=a+b-c is divided by U. But taking into account that the absolute value of the number a, b, c do not exceed number 0.5U, 4°) the number [math]h=a+b-c[/math] can have only three values: U, 0, -U. And now it follows from the equality 3° that 5°) the number [math]g=d+e+f[/math] can have respectively these three values: 0, U, 2U. And the following inequalities testify about truth of the FLT: I. If h=U and g=0, then [math] (U+t+0.5U)^n+(U+0.5U)^n-(2U+t+0)^n<max(H)=max(A^n+B^n-C^n)=[/math] [math]=(U+0.5U)^n+(U+0.5U)^n-(2U+0)^n<-1.24U<0[/math] with d=1, e=1, f=2; a=0.5U, b=0.5U, c=0 and any t>0. [With n=2 [math]max(H)=(U+0.5)^2+(U+0.5)^2-(2U+0)^2=0.5U>0[/math]!!! And equality H=0 is possible.] II. If h=0 and g=1, then [math] (U+t+0.25U)^n+(U+0.25U)^n-(U+t+0.5U)^n>min(H)=(U+0.25U)^n+[/math] [math]+(U+0.25U)^n-(U+0.5U)^n>0.53U>0[/math] with d=1, e=1, f=1; a=0.25U, b=0.25U, c=0.5U and any t>0. III. If h=-1 and g=2, than in this case either [math]B>1.5U[/math] (in the [math]H=(2U-0.5U)^n+(2U-0.5U)^n-(2U-0)^n)[/math], or [math]B<U[/math] (in the [math]H=(3U-0.5U)^n+(U-0.5U)^n-(2U-0)^n)[/math], that contradicts to 2°. (Equality – for example, a=0.5 or a=0, – it’s impossibly for that reason, which in this case, according to the theory of the Fermat’s equality, number A would have common divisors with the numbers B and C even after their division by the greatest common divisor.) Merged post follows: Consecutive posts merged 5°) the number [math]g=d+e-f[/math] can have respectively these three values: 0, U, 2U.

Proof of the Fermat’s Last Theorem (edited text) Let us assume that for integers A, B, C (C>A>B>0) and n>2 1°) [math] (H=) A^n+B^n-C^n=0[/math], where [math]A+B-C=U>0[/math] [is obvious, C>A>B>U] and 2°) [math]U<B<1,5U[/math]. Actually, from [math] [b^n=] BB^(n-1)=(C-A)Q[/math] it follows: [math]BB^(n-1)<(C-A)[nB^(n-1)][/math] (see formula for Q). From here [math]B<3(C-A), B<3(B-U), 3U<2B, B>1,5U[/math]. Let us accept for the base (for the unit of measurement) the number U: U=1. Now A=d*U+a*, B=e*U+b*, C=f*U+c*, or A=d*+a*, B=e*+b*, C=f*+c*, where a*, b*, c* are positive remainders after the division of the numbers A, B, C by U. If a<0.5U, then the remainder a* let us designate by letter a, and the number d* let us designate by letter d. If a*0.5U, then the number a*-U let us designate by letter a, and the number of d*+U let us designate by letter d. Analogously let us make also with the numbers e*, f*, b*, c*. As a result this number A, B, C is taken the form: A=d+a, B=e+b, C=f+c, where the positive or negative numbers a, b, c in terms of the absolute value less than 0.5. (Equality - for example, a=0.5 – is’t impossibly for that reason, which in this case, according to the theory of the Fermat’s equality, number A would have common divisors with the numbers B and C even after their division by the greatest common divisor.) From the equality U=A+B-C=(dU+a)+(eU+b)-(fU+c), or 3°) [math] (d+e+f)U+(a+b-c)=U[/math] follows that the number h=a+b-c is divided by U. But taking into account that the absolute value of the number a, b, c do not exceed number 0.5U, or 0.5, 4°) the number [math]h=a+b-c[/math] can have only three values: U, 0, -U, or 1, 0, -1. And now it follows from the equality 3° that 5°) the number [math]g=d+e+f[/math] can have respectively these three values: 0, U, 2U, or 0, 1, 2. And the following inequalities testify about truth of the FLT: I. If h=1 and g=0, then [math](1+t+0,5)^n+(1+0,5)^n-(2+t+0)^n<max(H)=max(A^n+B^n-C^n)=[/math] [math]=(1+0,5)^n+(1+0,5)^n-(2+0)^n<-1,24<0[/math] with d=1, e=1, f=2; a=0,5, b=0,5, c=0 and any t>0[/math]. II. If h=0 and g=1, then [math](1+t+0,25)^n+(1+0,25)^n-(1+t+0,5)^n>min(H)=(1+0,25)^n+[/math] [math]+(1+0,25)^n-(1+0,5)^n>0,53>0[/math] with d=1, e=1, f=1; a=0,25, b=0,25, c=0,5 and any t>0[/math]. III. If h=-1 and g=2, than in this case either [math]B>1.5U[/math] (in the [math]H=(2-0.5)^n+(2-0.5)^n-(2-0)^n)[/math], or [math]B<U[/math] (in the [math]H=(3-0.5)^n+(1-0.5)^n-(2-0)^n)[/math], that contradicts to 2°.

There is reason to believe that last lemma is incorrect. But to before end a study on FLT, I decided to rapidly examine the previous ideas, and in one of them (http://dxdy.ru/topic24331-15.html - Вт авг 11, 2009 16:28:24) I revealed the interesting passed moment: If AB not divides by n then the number q-p must be divided by n(a-b); however, in the field of real numbers (with prime n>2 and c>a>b>1) occurs the inequality n(a-b)>>q-p, which is magnificently confirmed by calculations on the computer: if A=15,96, B=12, C=17,9596, n=3, then the number q-p=0,722 are not divided by 3*(a-b)=1,659!!! And if A=125, B=64, C=130,353, n=3, then the number q-p=5,7096 are not divided by 3*(a-b)=6,9047!!! And it remains to us only confirm this fact by the approximate analytical calculations. [i will resemble: [math]A^n=C^n-B^n=(C-B)P=a^n*p^n; B^n=C^n-A^n=(C-A)Q=b^n*q^n[/math].] Merged post follows: Consecutive posts merged 1. Hypothesis is incorrect with the large numbers. 2. However, I found very interesting tool. Here is the project of new proof. Let us assume that for prime n>2 and integers A, B, C (C>A>B>0) 1°) [math]A^n+B^n=C^n[/math], where A+B-C=U>0. Let us designate by the letters A', B', C' remainder terms after division of the numbers A, B, C by U; by letters A", B", C" the number A'-U, B'-U, C'-U; by letters a, b, c the smallest (in terms of the absolute value) numbers in the pairs (A', A"), (B', B"), (C', C"). Thus, 2°) |a|<U/2, |b|<U/2, |c|<U/2 and 3°) A=a+dU, B=b+eU, C=c+fU (where d, e, f are integers). As it follows from 2°, the number v=a+b-с=U-TU can have only three values: U, 0, -U. Let us examine these cases. The I. a+b-с=U. Then (see 3°) d+e-f=0, and in this case 4°) [math] (dU)^n+(eU)^n-(fU)^n<0[/math]. However, addition to the numbers dU, eU, fU the numbers a, b, с with any values of them in the range of 2° CANNOT CHANGE this inequality: [math](dU+a)^n+(eU+b)^n-(fU+c)^n<0[/math]. The II. a+b-с=0. Then (see 3°) d+e-f=1, and in this case all the more 5°) [math](dU)^n+(eU)^n-(fU)^n<0[/math]. However, addition to the numbers dU, eU, fU the numbers a, b, с with any values of them in the range of 2° CANNOT CHANGE this inequality: [math](dU+a)^n+(eU+b)^n-(fU+c)^n<0[/math]. At last, The III. a+b-с=-U. Then (see 3°) d+e-f=2, and in this case all the more 6°) [math](dU)^n+(eU)^n-(fU)^n<0[/math]. However, addition to the numbers dU, eU, fU the numbers a, b, с with any values of them in the range of 2° CANNOT CHANGE this inequality: [math](dU+a)^n+(eU+b)^n-(fU+c)^n<0[/math]. Thus, the equality of 1° in the integers is impossible. Merged post follows: Consecutive posts mergedImpression, that the key to the proof of the FLT is found, did not change also after the second view. Now it is possible to approach the thorough calculation. However, now the process of proof is similar more greatly to the entertaining quiz: it suffices to know how to count only bis three. It is convenient to examine all numbers in the base U=1. Thus, the case I: a+b-с=U. Since in this case d+e-f=0 (see 3°) and (4°) [math](dU)^n+(eU)^n-(fU)^n<0[/math] (with minimum [math](1)^n+(1)^n-(2)^n<0[/math]), then it is necessary to maximize the number a and to minimize the number c. limiting values in this case: a=0.5, b=0.5, с=0. But since the minimum values of the numbers d, e, f are in any event equal to 1 (this it follows from A+B-C=U>0 and C>A>B>0), and the numbers a, b, c are not equal to zero (and to 0.5) and, in addition to this, d+e=f, then it is necessary to introduce correctives into the values of a=0.5, b=0.5, с=0. But even with these values of the numbers a, b, c the maximum value of the number [math]H=(dU+a)^n+(eU+b)^n-(fU+c)^n[/math] takes the form: [math]H=(1+0.5)^n+(1+0.5)^n-(2)^n[/math], and now even with n=3 (worst case) the value of the number H (=6.8-8) is nevertheless negative! And an increase in the numbers d and f by the equal number does not can to convert the number H from the negative into the positive. We pass now to the case OF THE II: a+b-с=0. Since in this case d+e-f=1 (see 3°) and in this case (5°) [math](dU)^n+(eU)^n-(fU)^n>0[/math] [with the minimum values [math](1)^n+(1)^n-(1)^n>0] [/math], then necessary to minimize the number a and to maximize the number c. The limiting values of the numbers a, b, c in this case are such: a=0.5, b=0, с=0.5. But even with these values a, b, c and d=e=f=1 the minimum value of the number [math]H=(dU+a)^n+(eU+b)^n-(fU+c)^n[/math] takes the form: [math]H=(1+0.5)^n+(1)^n-(1+0.5)^n[/math], and even this value of the number H is positive. Those more it will be positive and with any other values of the numbers a, b, c, d, e, f. I.e., with a+b-с=0 the number H is always positive. Finally, the case OF THE III: a+b-с=-U. Since in this case d+e-f=2 (see 3°) and in this case also (6°) [math](dU)^n+(eU)^n-(fU)^n>0[/math], then it is necessary to maximize on the module the numbers a and b and to minimize on the module the number c. The limiting values of the numbers a, b, c in this case are such: a=-0.5, b=0, с=0.5, and the numbers d, e, f are such: d=2, e=2, f=2. And the minimum value of the number [math]H=(dU+a)^n+(eU+b)^n-(fU+c)^n[/math] take the form: [math]H=(2-0.5)^n+(2)^n-(2-0.5)^n[/math], and even this value of the number H is positive. Those more it will be positive with any other values of the numbers a, b, c, d, e, f. I.e., with a+b-с=-U the number H is always positive. Thus, with any of the value of the numbers a, b, c the number H is not equal to zero. The FLT is proven. It is possible to approach the thorough checking of the proof. It is interesting that for this proof it suffices to have only general idea about the degree. Merged post follows: Consecutive posts mergedProof of Fermat's last theorem Let us assume that for the integers A, B, C (C>A>B>0) и n>2 1°) [math](H=) A^n+B^n-C^n=0[/math], where A+B-C=U>0. (From what it is evident that C>A>B>U.) Let us accept for the base (for the unit of measurement) the number U: U=1. Now A=d*U+x, B=e*U+y, C=f*U+z, or A=d*+x, B=e*+y, C=f*+z, where x, y, z are positive remainders from the division of the numbers A, B, C on U. If x<0.5U, then let us designate the remainder x by letter a and the number d* let us designate by letter d. If x>0.5U, then let us designate the number x-U by letter a and the number d*+U let us designate by letter d. Analogously let us to make also with the numbers e*, f*, y, z. As a result this the numbers A, B, C are taken the form: [math]A=d+a, B=e+b, C=f+c[/math], where the positive or negative numbers a, b, c in absolute value less than 0.5. It is evident from the equality [math]A+B-C=(dU+a)+(eU+b)-(fU+c)= [/math] 2°) [math]=(d+e+f)U+(a+b-c)=U[/math] that the number h=a+b-c is divided into U. But taking into account that in absolute value of the numbers a, b, c do not exceed number 0.5U, or 0.5, then 3°) the number [math]h=a+b-c[/math] can have only three values: U, 0, -U, or 1, 0, -1. And now it is evident from the equality 2° that 4°) the number [math]g=d+e+f[/math] can have only these three values: 0, U, 2U, or 0, 1, 2. Finally, for these three pairs of values h and g we can write down three inequalities: The I. If h=1, then g=0 and with the complete obviousness [math]max(H)=max(A^n+B^n-C^n)=max[(d+a)^n+(e+b^n-(f+c)^n]= [/math] [math]=(1+0.5)^n+(1+0.5)^n-(2+0)^n<0[/math] (here d=1, e=1, f=2; a=0.5, b=0.5, c=0). The II. If h=0, then g=1 and with the complete obviousness [math]min(H)=min[(d+a)^n+(e+b^n-(f+c)^n]=(1+0.5)^n+(1+0)^n-(1+0.5)^n>0[/math] (здесь d=1, e=1, f=1; a=0.5, b=0, c=0.5). The III. If h=-1, then g=2 and with the complete obviousness [math]max(H)=[/math] [math]=(2-0.5)^n+(2-0.5)^n-(2-0)^n<0[/math] (here d=2, e=2, f=2; a=0.5, b=0.5, c=0). With any changes in the numbers a, b, c, d, e, f, but, of course, with the observance of the conditions of 1°, 3°, 4°, these inequalities can only BE STRENGTHENED. Thus, the number H, or [math]A^n+B^n-C^n[/math], IS NOT EQUAL to ZERO Fermat's last theorem is proven fully. Merged post follows: Consecutive posts mergedP.S. Possibly, in the proof of the point of 5° it is necessary to use the simple lemma: U<B<2U. This means that e*=1; consequently, with b>0 e=1, and with b<0 e=2.

Yes. Thank. ++++++++++ Last idea required refinement, the following lemma became result of which: If for the integers a, b, c, p, q, r is the equality ap+bq-cr=abc>0 (c>a>b), then p+bq-cr>0. Is there a simple proof of this lemma? + + + If lemma is true, then Fermat's last theorem proves by one phrase. Are actual, with relatively prime [math]A, B, C[/math] 1°) [math]A^n=C^n-B^n=(C-B)P=a^n*p^n[/math] [or 1b°) [math]B^n=C^n-A^n=(C-A)Q=b^n*q^n] [/math], where, according to lemma, 2°) [math]p>a[/math] (or [math]q>b[/math]). And now during the term-by-term multiplication of Fermat's equality by the sufficiently large number [math]d^{nn}[/math] 3°) the number [math]ad^n[/math] becomes more than the number [math]pd^{n-1}[/math], that contradicts 2°.

Fantastic fundamental contradiction of Fermat’s equality It is completely obvious that for relatively prime natural A and B in the number 1°) D=a^n+b^n=(a+b)R occurs the inequality: a+b<R. However, if D=c^n, then inequality sign changes by the opposite. The simple proof of this fact will be represented later. ========== Prompt for the too impatient: R<c^n-(c-1)^n<(c^n)^(0,5) (you will recall the formula of the expansion of the sum of degrees). Consequently, a+b>(c-1)^n>(c^n)^(0,5). From where (a+b)>R, that contradicts 1°.

C-B and P have no common factors, because the number P can be represented in the form: P=T(C^2-B^2)+n(CB)^(n-1). Merged post follows: Consecutive posts merged Yes! Thank You very much!

If the numbers C and B have no common factors and C-B divided not by n, then the numbers C-B and P=(C^n-B^n)/(C-B) have no common factors. Therefore if a' is a prime divider of the number C-B, then a'^n is a divider of the number C-B and is not a divider of the number P. a is a designation of the product of all prime dividers a' of the number C-B, b is a designation of the product of all prime dividers b' of the number C-A, c is a designation of the product of all prime dividers c' of the number A+B. Victor Merged post follows: Consecutive posts mergedStrictly speaking, the essence of the represented proof of the FLT consists in all of the most primitive linear equation of 5°-11° with one unknown x and the interpretation of its solution. The lines of 1°-4° - these are well well-known facts, and the discrepancy of the formula of 11° can be shown by many methods. Here one more. The number [(C-B)+(C-A)]=c[2r-c^(n-1)] in 11° it is divided only into c (since it is not divided by r), i.e., 12°) [(C-B)+(C-A)]=c. But with the aid of the reasonings of 1°-11° we can obtain analogous equalities, also, for the numbers A and B: 13°) [(A+B)-(C-B)]=b, 14°) [(A+B)-(C-A)]=a. And if A+B-C=U, where, as is known, U is divided into abc, then, adding the equalities 13° and 14° and reading from the sum the equality 12°, we have: 15°) 2[(A+B)-(C-B)-(C-A)]=4(A+B-C)=4U=a+b-c. Consequently, the number a+b-c is divided completely into 4abc, which is, obviously, impossible for natural a, b, c.

Last nail Proof of Fermat's last theorem for the prime n>2 Let us assume that for the relatively prime natural a, b, c there is a equality: 1°) A^n+B^n=C^n. Let for the certainty the number A and B be not multiple n. In this case, as is well known, there are the following relationships and the ideas: 2°) A^n=(C-B)P, B^n=(C-A)Q, C^n=(A+B)R, A^n-B^n=(A-B)T; the number in the sets 3°) [A, B, C], [C-B, C-A, A+B, A-B], [P, Q, R, T], [R, A+B, if A+B not is multiple n] relatively prime, but if A+B is multiple n, then R is multiple n and not multiply n^2; 4°) C-B=a^n, C-A=b^n, A-B=a^n-b^n, A+B= either c^n (if A+B not is multiple n), or (c^n)/n (if A+B multiply n). Proof of the FLT We have from the equalities A^n-B^n=(A-B)T=(C-B)T-(C-A)T and A^n-B^n=(C-B)P-(C-A)Q: 5°) (C-B)T-(C-A)T=(C-B)P-(C-A)Q, from where 6°) (C-B)(P-T)=(C-A)(Q-T). Since the number C-A and C-B have no common factors, then the number P-T is divided by C-A, and the number Q-T is divided into C-B, i.e., we can write down: 7°) P-T=(C-A)x, Q-T=(C-B)y. Let us substitute these values into 6°: 8°) (C-B)(C-A)x=(C-A)(C-B)y, from where x=y. And now we have from 7°: 9°) P=T+(C-A)x, Q=T+(C-B)x. Let us substitute these values into the equality A^n+B^n=(C-B)P+(C-A)Q: 10°) A^n+B^n=(C-B)P+(C-A)Q=(C-B)[T+(C-A)x]+(C-A)[T+(C-B)x]= =[(C-B)+(C-A)]T+[(C-A)+(C-B)]x= 11°) =[(C-B)+(C-A)](T+x)=C^n. It is easy to see that the number 12°) D=(C-B)+(C-A)=2C-(A+B)=2cr-c^n=c[2r-c^(n-1)] contains only one divider - the number c, the number D is not divided into the number r, and the number C does not contain dividers beyond the limits of the dividers of the number cr. However, D=2C-(A+B)>c, since for C>n 13°) D=(C-B)+(C-A)>[nC^(n-1)]^(1/n)+[nC^(n-1)]^(1/n)=2[nC^(n-1)]^(1/n)>c. But if C is multiple n, then the relationship 13° those more is fulfilled, since the number T not is multiple n, but the number x is multiple n.

For me it was possible to lead Fermat’s equality for prime n to the very interesting form: either: A=pU+a, B=U+b, C=pU+c, where A, B, C is natural relatively prime, U=A+B-C and multiple by n^2, a+b-c=0, 0<(a, b, c)<U; among them there are no equal, among the numbers a, b, c two are not deliberately multiple n, or: A=pU+e, B=U+d-e, C=pU+d, where U>d>e>0: (pU+e)^n+(U+d-e)^n=(pU+d)^n. The equality a+b-c=U, as it is very easy to show, under the conditions FLT is impossible. Possibly, so easily it will be possible to prove the impossibility of the equality of a+b-c=0.

Two days without the Internet *** Proof of FLT by P.Ferma (hypothesis) Continuation after 8°: Let us multiply the equality 1° by this number D= (d+ut)^n, that in new equality 9°) U>c'^n. If now 10°) a+b-c'=0, the last figures in the base U in the equality of 1° give the non-integral equality of a^n+b^n-c' ^n. But if 11°) a+b-c' =U, then after the substitution a=U-a'' (or b=U-b''), where now a''+b-c'=0, last digits in the base U in the equality 1° give the non-integral equality 12°) (-a'')^n+b^n-c'^n=0 (or a^n+(-b'')^n-c'^n=0'), that with entire a'', b, c' it is impossible. Great theorem is proven. Merged post follows: Consecutive posts mergedLetter to the forum Gentlemen, The second proof of FLT consists, actually, of one operation of the multiplication of Fermat’s equality by the large number D, after which we immediately obtain conclusion about the discrepancy of equality. Try to understand my proof, and I will answer all your questions. I hope that the imperfect transfer will not become obstacle to the mutual understanding. Victor Sorokin Merged post follows: Consecutive posts mergedAnswer to questions of Moshe Klein 1) What is wrong in this prove for the case n=2? - Cf.4°: 4°) c' (>1) - the greatest common divisor of the numbers C and U (C= c'c*, U=uc'). If n=2, then always c' =1; If prime n>2, then always c' >1, since A^n+B^n=(A+B)R, where the greatest common divisor of the numbers A+B and R always more then 1. 2) What is the improvement from your last version of FLT? - Both last versions of proof are accurate, but the second version is considerably shorter and it is simpler. 3) Why you say by Ferma? - Because 80% of proof this of FLT is a formula of the general solution of the linear diophantine equation 8°) c*(d+ut)-u(e+c*t)=1. Merged post follows: Consecutive posts merged Corr.: 1) What is wrong in this prove for the case n=2? - Cf.4°: 4°) c' (>1) - the greatest common divisor of the numbers C and U (C= c'c*, U=uc'). If n=2, then always c' =1; If prime n>2, then always c' >1, since A^n+B^n=(A+B)R, where the greatest common divisor of the numbers A+B and U always more then 1. Merged post follows: Consecutive posts merged Analysis of proofs. 1. Both versions of proof are based on the well-known fact: the numbers C and U contain the dividers c*>1 and u>1 not being been dividers numbers A+B; the numbers c* and u are interprime (relatively prime). (With n=2 this condition is not satisfied. Hence the hypothesis: with relatively prime A, B, C the number C in the equality A^n+B^n=C^n is prime.) 2. Positive integer relatively prime numbers c* and u generate the linear diophantine equation c*(d+ut)-u^n(e+c*t)=1. 3. After the multiplication of Fermat’s equality on D=(d+ut)^n with sufficiently large parameter t we obtain equivalent Fermat’s equality A''^n+B''^n-C''^n=0, where A''=a''+p''U'', B=b''+q''U'', C''=c''+r''U'' (где A''+B''-C''=U''), in which on the base U'' the number c'' is equal to the number c' from the previous equality, since C(d+ut)=c'c*(d+ut)=c'[c*(d+ut)]=c'[c*(d+ut)]=c'[u^n(e+c*t)+1]=c'. I.e. on the infinite set of the numbers D=(d+ut)^n (t=1, 2, …) the number c' is CONSTANT. 4. And now we see that both last versions of proof of the FLT ВТФ are accurate: in the first, A''^n+B''^n-C''^n>0; in the secondly, a''^n+b''^n-c'^n=0 (with a''+b''-c'=0) or a''^n-b''^n-c'^n=0 (with a''-b''-c'=0). Problem of Fermat in the unofficial science is exhausted.

*** Elementary proof of FLT Let us assume that for natural A, B, C 1°) A^n+B^n=C^n [or E=A^n+B^n-C^n=0], where prime n>2, the number 2°) A+B-C=U, C>A>B>U>0 and, therefore, 3°) C-U>A-U>B-U>0; 4°) c' (>1) - the greatest common divisor of the numbers C and U; C= c'c*, U=uc'. Let us write down the numbers A, B, C on the module U: 5°) A=a+pU, B=b+qU, C=c+rU (here a, b, c are positive number-digits, <U). 6°) if c is not equal of c', then let us multiply the equality of 1° by the sufficiently large number D= (d+ut) ^n, where the number (d+ut) is undertaken from the solution of the linear diophantine equation 7°) c*(d+ut)-u(e+c*t)=1. About the value of the number (d+ut)^n will be said below. 8°) But is important that the new value of the number C in 5° (WITH ANY (d+ut)^n!) will take the form: C=c'+rU, where c'<<U and it is constant by the change t. Possibly, 6°-8° are superfluous, nevertheless we will examine the infinite sequence of Ferman’s equalitie 1°, multiplied by D= (d+ut)^n (t=1,2,…). It is easy to see that from 5° (taking into account 2° and 3°) escape two possibilities: 9°) a+b-c=0, and then pU+qU-rU=U, or 10°) a+b-c=U, and then pU+qU-rU=0. 11°) It is important that in both cases q=1 (otherwise, taking into account 2°, as the module one should take the number qU.) Let us substitute 5° into 1°, let us discover Newton's binomials and let us group the similar terms: 12°) E=(a+pU)^n+(b+qU)^n-(c'+rU)^n=0, or E=U^n(p^n+1-r^n)+nU^{n-1}(ap^{n-1}+b^{n-1}-c'r^{n-1})+…+ + nU(a^{n-1}+b^{n-1}-c'^{n-1})+(a^n+b^n-c'^n)=0. Let us show now that on the infinite set of multipliers D=(d+ut)^n (see 6°-8°) for the equality 1° the number E>0. Case of 9° (taking into account of 8° and 11°): a+b-c' =0 and pU+U-ru =U, where p=r, U>u>c' >a>b>0. Let us discover Newton's binomials and it will group the similar terms: E=(pU+a)^n+(U+b)^n-(pU+c')^n= =(pU)^n+U^n-(pU)^n+ n[(pU)^(n-1)a+U^(n-1)b-(pU)^(n-1)c']+ … … +n[a^(n-1)(pU)+b^(n-1)U-c'^(n-1)pU]+(a^n+b^n-c'^n). Let us take the now reduced value of the number E: 13°) E'=U^n- n(pU)^(n-1)c'- … -nc'^(n-1)pU-(c'^n-a^n-b^n), where 0 0<c'^n-a^n-b^n<nc'^(n-1)pU (taking into account that c'-a-b=0). Let us decrease now and this value of E': let us first, increase the last term of c'^n-a^n-b^n to the value of nc'^(n-1)pU, in the second place, except the first, let us increase all binomial coefficients to maximum T (in the binomial of the n-th degree), thirdly, the exponents with the bases pU and c' in all terms let us increase to n–1. As a result the considerably reduced value of E' it will become equal 14°) E*=U^n-nT(pc')^(n-1)U^(n-1)=U^n-VU^(n-1), where the V is constant. And having now taken in 6° the sufficiently large number (d+ut)^n as the coefficient of the equality of 1°, we can obtain the value of the number E*, therefore, and the numbers E' and E, greater than any given number. Which was required to prove. Case of 10° (taking into account of 8° and 11°): a+b-c' =U and pU+U-rU=0, where p+1=r This case is reduced to previous with the aid of the substitution, achieved only in the last term of expansion E into 12° (since the sum of all rest it takes the form V'U^(n -1), where V' is constant): 15°) a=U-a'', b=U-b'', c'=U-c'', where on the infinite set of the coefficients (d+ut)^n (see 6°-8°) for the equality of 1° the digit c'', consequently and the number c''^n, they are constants, and the number a''+b''-c''=0. And now (just as in the preceding case), having substantially decreased the number E into 12° – after replacing positive terms, besides U^n and c''^n, with zeros, and each negative item, except a''^n and b''^n, decreasing to -T(pc')^(n-1)U^(n-1), and decreasing the number -(a''^n+b''^n-c''^n) to -nT'(c'')^(2n-2)U^(n-1), – we obtain the negative number E*=U^n-nT*(pc')^(n-1)U^(n-1)-nT'(c'')^(2n-2)U^(n-1)= U^n-VU^(n-1), which with the sufficiently high coefficient D=(d+ut)^n (см. 6°-8°) for the equality 1° exceeds any given number (i.e. just as in the first case). The negative number E* after its multiplication by the positive D=(d+ut)^n becomes the number positive, which is impossible. Thus, Fermat's last theorum is completely proven. Merged post follows: Consecutive posts merged The many thanks! But 1) I do not speak in English, 2) if the sense appears... Victor

If we will take multiplier d from the general solution of the linear diophantine equation c*(d+ut)-u(e+c*t)=1, where C* and u is the greatest and relatively prime dividers of the numbers C and U and C=c'c*: d = d+u, d+2u, d+3u, …, then C=[u(e+c*t)+1]c'=TU+c', where c' is CONSTANT! So: U^n>>>T*U^(n-1)+c'^n!

If A=a+pU, B=b+qU, C=c+rU (A+B-C=U), then (after the multiplication of Fermat’s equality by the sufficiently large number d^n) A^n+B^n>C^n!!! In all cases! Proof is transferred in the English. Thanks Merged post follows: Consecutive posts merged++++++++ If A^n+B^n-C^n=0, then, after the substitution A=a+pU, B=b+qU, C=c+rU (where U=A+B-C), the number A^n+B^n-C^n has the form: U^n-VU^(n-1), where V is constant. And after the multiplication of Fermat’s equality by the sufficiently large number d^n the number A^n+B^n>C^n!!! In all cases! Proof is transferred in the English.

Yes! Thank! Merged post follows: Consecutive posts merged More precise calculations give identical results. Theme is closed. Sorry...

1) If: a=16,43; b=11,82; c=18,25; a+b-c=u=10, then a^3+b^3-c^3= =4435,19+1651,40-6086,6=0.

In the first case we find the value of the number a^3+b^3-c^3 WITHOUT the use of Newton's binomial, in the second case - with the use.

My proof is wrong! BUT... ++++++++++++++++++++++ Fairy phenomenon Foundations of the mathematics incorrect! Example: 1) If: a=16,43; b=11,82; c=18,25; a+b-c=u=10, then a^3+b^3-c^3=0. But if: a=(10+6,43); b=(10+1,82); c=(10+8,25); a+b-c=10, then a^3+b^3-c^3=8,06! *** Calculation: [1000+3*10*(41,34+3,31-68,06=-23,41)+(265,85+6,03-561,52=-289,64)= =1000-702,3-289,64=8,06] Thus: 0=8,06!!! This phenomenon is true for ANY positives a, b, c! (There is the Proof.) Еще один пример: 2) If: a=19,5013; b=11,1433; c=20,6446; u=10; then a^3+b^3-c^3=0. But if: a=(10+9,5013); b=(10+1,1433); c=(20+6446; u=10); then a^3+b^3-c^3=-23,23. Thus: 0=-23,23!!! Calculation: [-6000+3*(100*9,5013+100*1,1433-400*0,6446=806,62)+ -3*(10*90,269+10*1,3071-40*0,4155=899,141)+(857,727+1,4944-0,2678=-326,98) = = -6000+2419,86+2697,423+859,489=-23,23] Thanks!

Correction of the error Fermat's last theorem The designations of the numbers, which participate in the proof, become clear from the following relationships after the elimination of common factors in the numbers A, B, C (let us examine only the case, when AB not divided by simple n): 1°) A^n+B^n=C^n и 1a°) A^n=C^n-B^n=(C-B)P=(a^n)(p^n)=(ap)^n, 1b°) B^n=C^n-A^n=(C-A)Q=(b^n)(q^n)=(bq)^n, 1c°) C^n=A^n+B^n=(A+B)R. 2°) It is important that the numbers A-B, C, a, are relatively prime! Classical proof of FLT Let us examine the number D=A(C-B)-B(C-A). From one side, 3°) D=A(C-B)-B(C-A)=(A-B)C. But from the other side, 4°) D=A(C-B)-B(C-A)=Ab^n-Ba^n=apb^n-bqa^n=ab(pb^{n-1}-qa^{n-1}). And comparing 3° from 4°, we see that the number (A-B)C is divided by ab, that contradicts to 2°. FLT is proven. 23.09.2009

Elementary proof of the FLT The designations of the numbers, which participate in the proof, become clear from the following relationships after the elimination of the general cofactors 0°) - let us assume that this is possible, - in the numbers A^n, B^n, C^n (for the beginning AB is’not divisible by n): 1°) A^n+B^n=C^n, 2°) A^n=C^n-B^n=(C-B)P=a^np^n=(ap)^n, 3°) B^n=C^n-A^n=(C-A)Q=b^nq^n=(bq)^n, 4°) C^n=A^n+B^n=(A+B)R, 5°) A+B-C=U=(A+B)-C=A+(B-C)=B+(A-C)=ap-a^n=bq-b^n=uabc*, where c* - greatest common divisor of the numbers C and A+B (it’s obviously that c*>1), from here 6°) C=A+B-uabc*. 7a°) It’s obviously that the numbers A^n+B^n, C^n-B^n, C^n-A^n, A^n-B^n, C^n+B^n, C^n+A^n are in pairs relatively prime. 7°) Obvious that the numbers A+B, C-B, C-A, A-B, C+B, C+A are also relatively prime. Proof Let us examine two numbers: C-A and C+A, or, taking into account 6°, 8°) C-A=B-uabc* and C+A=2A+B-uabc*. And now it is easy to see that also the sum [2A+2B-2uabc*, or 2(A+B-uabc*)], and the difference [2a] of the numbers C-A and C+A ARE DIVIDED into c* (see 6° and 5°), that contradicts to 7°, 7a°, 0°. If the number CB is not divisible by n, than the proof is perfectly analogy. FLT is proven.

http://www.scienceforums.net/ is the only forum, where it is possible to publish ideas from physics. Merged post follows: Consecutive posts merged It is not possible to measure the rotational energy with the aid of the kinetic thermometer. Merged post follows: Consecutive posts merged 1. It is not possible to measure the rotational energy with the aid of the kinetic thermometer. 2. Interest? It is today known 100 types of turbines. In 1986 I found 1000000 (one million = all!!!) new forms and created the complete concept of turbine. And no interest!

Rewards for the turbines and the converter of rotational energy of the gas: Silver medal, Geneva, 1988; Gold medal even two certificates, Lyon, 1998. Merged post follows: Consecutive posts merged Answer is here: HOW TO USE THE ROTATIONAL THERMAL ENERGY OF GAS Rotational Thermal Energy (or RTE) was already found in the molecular theory of gas by Boltzmann. So far, however, nobody has demonstrated yet how it could be used. This work is the first attempt to do so. 1. In the heat-engines the RTE of molecules is not transformed into mechanical energy, because the piston (and the turbine) moves by the force, which is oriented the same direction as its motion. 2. Therefore, the RTE of the gas does not change when a mechanical work is done. 3. Because the surface of the thermometer also does not take in the tangential effort, all figures showing the calorific values of the fuel are less by the value of the RTE. 4. As follows from the mechanics, a molecule of gas created during a chemical reaction gets a high RTE. A large portion of kinetic and vibrational thermal energy (KVTE) of constituent elements and the work of the force of chemical attraction when the molecules are close to each other are transformed into RTE. 5. If one of the elements of that reaction is a solid and immobile substance, then the RTE of the molecule of gas is created only by the mobile element, and the KVTE is created only by immobile element. And it is proportionally to the masses of these elements. So, when crystalline carbon is chemically reacting with oxygen and CO2 is formed, the proportion RTE/KVTE = 16:6 = 2.667, the fact which is confirmed experimentally up to four digits precision. 6. No phenomenon of self-heating of exhaust gas (i.e. spontaneous transformation of RTE into KVTE) is observed. Hence, all RTE of combustion product is totally wasted. 7. The fact of existence of thermal energy of unknown origin has been experimentally observed in 1818: a "catalyst" (tiff) raises the calorific value of the coal by 2.668 times. 8. Almost complete coincidence of the theoretical value of the RTE of carbonic acid (gas) with the experimental one gives us total confidence to think that the "catalyst" is indeed the transformer of RTE into KVTE, or RTE-transformer. The above-mentioned facts allow us to explain the machinery of the RTE-transformer: after a chemical combination of a molecule of CO2 with an immobile molecule of CaO the rotation of the molecule of CO2 stops and its RTE is transformed into KVTE. However, the value of KVTE turns out to be so high that it provokes a disintegration of CaCO3 back to CaO, but immobile, and a CO2, but not rotating. The model of RTE presented here fully explains all cases of appearance of a mysterious extra thermal energy in many thermodynamic processes and allows us to find RTE-transformers with the highest efficiency. This means that it is possible to raise the calorific value of chemical fuel in all existing appliances including heat engines. Thus, the presented model of RTE raises by 2 to 3 times the potential of the thermal energy available to the mankind. It should be noted that physical and chemical properties of gas with high RTE substantially differ from that of the ordinary gas. Victor Sorokine (France. 1998) Merged post follows: Consecutive posts merged No one buys idea - they steal them.

In 1985-86 years I made a number of discoveries and several hundred inventions in the field of savings and production of super-cheap alternative energy. About hundred inventions I even patented or made a patent application. Among them: super-cheap in the production slow hydro-turbine with efficiency more than 98%, turbine with constant efficiency, turbine engine with efficiency 56%, steam engine with efficiency to 72% (today their efficiency it does not exceed 7%), the increase of the caloricity of chemical fuel en 2 and more times (that is equivalent to the doubling of world's supplies of coal and oil and gas), super-cheap variable-speed drive (automatic gearbox with a continuous change in their values), the reduction of energy losses tn 3-10 times, etc. For the payment of other patents I did not have money. Two-year attempts to realize my inventions showed the futility of this undertaking: Sale of inventions requires the high financial expenditures, which i did not have. And therefore I proposed my discoveries and inventions into the gift to each liberal-democratic state and I wrote the letter about my proposal. However, the super-cheap methods of obtaining the energy do not still have a demand… In the list of the proposed inventions also such were indicated: super-cheap (into hundred times!) heliostation with the payback period only 3 (!) month, the super-cheap (into thousand of times) storage batteries of energy for millions kilowatt-hours, two-cycle internal combustion engines with efficiency 56% (today their efficiency is equal to 18%), wave power stations of direct transformation of the energy, heat engines, which work only on the water, an increase in the caloricity of the chemical fuel, which goes to the heating of accomodations, into 100 times, super-cheap rocket fuel, etc. Are since then past twenty years. Energy crisis in the world began, but super-cheap energy does not have a demand. This strange nevertheless civilization… =========== The USA can ensure the entire world with energy.

It seems, we arrived! Here is actually the complete text of the proof of the FLT. In the prime base n>2 in the equality (1°) a^n+b^n-c^n=0 the number (sum of digits) (2°) a+b-c=U=un^k, where k>0 and u not it is divided by n. But then (3°) U+0=(a+b-c)+a^n+b^n-c^n=a(1+a^{n -1}) +b(1+b^{n -1})-c(1+c^{n -1})=U. But the equality 3° is contradictorily, since if abc not is divided by n, then the last non-zero digit in the number U has simultaneously TWO different values: in the number U+0 it is doubly (since, taking into account to Fermat's litle theorem, all three expressions in the parentheses finish are finished by digit 2) more than in the number U (but if the doubled value exceeds n, then from it should be deducted base n). But if, for example, the number of c=c'n^t, where c' not is multiple n and t>0, then in the new value of the number U [t (n -1) of +1]-th (from the end) digit in the number c will increase by 1, and tn-th digit in the number a+b will be doubled. Here, strictly, and everything. 2008.10.10 But this is how appears brief proof of the FLT (for the basic case: abc not divided by ghbme n> 2): In the Fermat’s equality the number u=a+b-c IS NOT EQUAL to the number U+0=(a+b-c)+a^n+b^n-c^n=a(1+a^{n-1})+b(1+b^{n-1})-c(1+c^{n-1}), since in the second number all expressions in the parentheses finish, according to Fermat's Little theorem by digit 2 and, therefore, the last significant (non-zero) digit of the number U+0 is twice more than as the last significant digit in the number U. In order to refute my proof, it is necessary to prove that: The multiplication of single-digit positive number by 2 does not change its value. But to whom this on the teeth?