Victor Sorokine

Clean proof of Fermat's last theorem “I found the truly fairytale proof of this remarkable theorem, but, unfortunately, place in the fields it is insufficient in order to give it here” /P.Ferma/. For lack of time today I will describe only about method and essence of proof. In the essence school proof consists of the following: In the equality [math]a^n=c^n-b^n=(c-b)P=AP[/math] the remainder from the division of the number P into A, calculated BY TWO methods, gives two DIFFERENT values. * * * Utilized mathematical tools: formula for the polynomial P and the square of a difference of two numbers. And all. Continuation follows.

In what joke? Complete proof of Fermat's last theorem Let us assume that for relatively prime natural [math] a[/math], [math]b[/math], [math]c[/math] ([math]c>a>b[/math]) and [math]n>2[/math] there is a equality: 1°) [math]a^n+b^n=c^n[/math], where, as is known, [math]a-b>1[/math]. Then the number 2°) [math]D=c^n+a^n=a^n+b^n+a^n=2a^n+b^n[/math], i.e. [math]2a^n+b^n[/math] is the sum of two degrees: [math]c^n+a^n[/math]. However, it is completely obvious that not be can, since the number [math]b^n[/math] is n-th degree with any b, while the number [math]2a^n[/math] is not n-th degree on no account. Thus, the equality 1° has no solution in integers. FLT is proven. ============== To show the inapplicability of proof in the case n=2 is given to the readers. I will indicate only that in this case the number a-b=1 and Fermat’s equality is reduced to the form: [math]2ab+1=c^2[/math].

Fermat's last theorem. Elementary proof Joke of the amateur [The key of the proof: If A+B=1, where A and B are integers and non-negative numbers, then either A or B is equal to zero] Proof of the FLT If for mutually-prime integers A, B, C and prime n>2 there is an equality 1°) [math]A^n+B^n=C^n[/math], where [math]C^n=(A+B)R[/math] or (if C divided by n) [math]C^n=(A+B)Rn[/math], where, as is known, 2°) the number A+B and R are relatively prime, then there is a positive solution (x, y) of linear diophantine equation 3°) [math] (A+B)x-Ry=1[/math]. It follows from this solution that in the numeration system on the base R the number 4°) A+B finishes by digit 1. However, since the numbers A and B are wholes and non-negative, the either number A or the number B finishes by zero, i.e., it is divided by R, which contradicts to condition. The case of n=4 proves analogously with the equation C^4-A^4=B^4. Theorem is proven. The case of n=4 proves analogously with the equation [math]C^4-A^4=B^4[/math]. Theorem is proven

Small correction It is insufficient for the proof of the FLT one Theorem alone, since the case of prime r (as the case of prime c in the equality [math]A+B=c^n[/math]) does not prove FLT. However, if we instead of r take the number C=cr, then on the basis Theorems we obtain the conclusion that the number C=cr is prime. Which is impossible. Thank to ALL

Theorem* (and FLT) for “Dummies” Theorem is true for composite r (that is sufficient for the simplest the proof of FLT). This evidently at least based on the example of decimal system: Numbers 7, 9 and 10 are relatively prime, and the coefficient d=5 (to prime divider of base). And now the numbers of 7x5 and 9x5, equal to 35 and 45, finish TO one and the same digit 5. (Understandable that for the proof of FLT as r it is necessary to take, for example, rr.) 400 years were required, in order to formulate this theorem!!! ____________ Theorem* If integers a, b, a+b and r are mutually-prime, then there is such d, relatively prime with r, that the ends of the numbers ad and bd are equal on the module r. [Consequence. With relatively prime a, b, a+b and r, where the value r is undertaken from the equality: 1*) [math]a^n+b^n=(a+b)r^n=c^n[/math] or [math]a^n+b^n=(a+b)nr^n=c^n[/math], equality 1* is contradictory in the base r, since in the equality 2*) [math](ad)^n+(bd)^n=(cd)^n[/math] ([math]=Pr[/math]) right side is divided by r, but leftist is not divided.]

Because of the misunderstanding my post is published in another theme: http://www.scienceforums.net/topic/50825-puzzle-on-the-course-theory-of-the-numbers/ I please apology.

Puzzle on the course “Theory of the numbers” Theorem If integers a, b, a+b and r are mutually-prime, then there is such d, relatively prime with r, that the ends of the numbers ad and bd are equal on the module r. *** [Consequence. With relatively prime a, b, a+b and r, where the value r is undertaken from the equality: 1*) [math]a^n+b^n=(a+b)r^n=c^n[/math] or [math]a^n+b^n=(a+b)nr^n=c^n[/math], equality 1* is contradictory in the base r, since in the equality 2*) [math](ad)^n+(bd)^n=(cd)^n[/math] ([math]=Pr[/math]) right side is divided by r, but leftist is not divided.] Proof is located in the stage of formulation.

Proof of Fermat's last theorem for relatively prime A, B, C and prime n>2. (Most splendid joke) Proof is based on the simple consequence from the Little fermat's theorem: In the numeration system on the base n the number R in the equality [math]A^n+B^n=(A+B)R[/math], where A, B and A+B are not multiple n, finishes with number 1. Case 1: ABC not is multiple n As is known, in this case in the Fermat’s equality [math]A^n+B^n-C^n=0[/math], or 1°) [math](C-B)P+(C-A)Q-(A+B)R=0[/math], the following formulas occur: 2°) [math]C-B=a^n, P= p^n; C-A=b^n, Q=q^n; A+B=c^n, R=r^n[/math]. 3°) In this case the numbers P, Q, R finish with 01. Proof of case 1 Let us examine the equality of 1° at the two-digit ends of the numbers (i.e. on the module [math]n^2[/math]). After rejecting coefficients P, Q, R as being finished with 01, we obtain the equation 4°) [math](C-B)+(C-A)-(A+B)=0[/math], or [math]a^n+b^n-c^n=0[/math] (on the module of [math]n^2[/math]), which also can be represented in the form 1°: 1_1°) [math](C_1-B_1)P_1+(C_1-A_1)Q_1-(A_1+B_1)R_1=0[/math], where 2_1°) [math]C_1-B_1=a_1^n, P_1= p_1^n; C_1-A_1=b_1^n, Q_1=q_1^n; A_1+B_1=c_1^n, R_1=r_1^n[/math]. 3_1°) in this case the numbers [math]P_1, Q_1, R_1[/math] finish with 01. And so on. Obviously, if the numbers A, B, C contain one divider, which is finished to the number, different from 1, then the process of release from the cofactors, which are finished with 01, will be INFINITE. This means that the numbers A, B, C are infinite. Case of 2, when A (either B or C) is multiple of degree n, proves with the aid of the same apparatus, but somewhat more complex.

On one incomplete proof of the FLT There is a very simple and brief (5 lines) proof of the FLT, based on the following theorem for the exponential binomials: Theorem For the relatively prime numbers A and B and A+B not divided by prime n>2 each prime divider m of the number R in the identity [math]A^n+B^n=(A+B)R[/math], where [math]A=a^{n^k}[/math] and [math]b=b^{n^k}[/math], but number a or/and b are not n-th degrees, each prime divider m of the number R is represented in the form [math]m=dn^{k+1}+1[/math], where d not is divided by n. *** Unfortunately, i do not have complete proof of this theorem. With the aid of the formula of the solution linear diophantine equation I found the proof only of that fact that the number R does not contain the dividers of the form of [math]m=dn^k+1[/math]. It will be possible to prove that R does not contain the dividers of the form of [math]m=dn^{k+2}+1[/math] and so forth? But can, it will be possible to prove FLT without this part… Only the base case is thus far until today proven: The number R (either P or Q) contains at least one divider [math]m=dn+1[/math]. (This, in particular, means that now the number A+B-C is divided by [math]n^3[/math].)

Proof of FLT for case 2 (material for the deep reflections) Let us assume for relatively integers A, B, C and prime [math]n>2[/math] the equality 1°) [math]C^n-A^n-B^n=0[/math] is possible and let the number C [either A or B] finishes on k zeros. Then the numbers [math]C^n[/math] and [math]A^n+B^n[/math] [[math]=(A+B)R[/math]] are divided by [math]n^{kn}[/math], 2°) R is divided by [math]n^1[/math] and not divided by [math]n^2[/math] (well-known fact), therefore, 3°) A+B is divided by [math]n^{kn-1}[/math], 4°) the number [math]A+B-C [=U][/math] is divided by [math]n^k[/math]. Proof of the FLT Let us examine the number [math]U^n[/math], or [math]U^n+0[/math] 5°) [math]U^n+C^n-(A^n+B^n)=(U+C)T-(A^n+B^n)[/math], where [math]U+C=A+B[/math] (see 4°) and, therefore, is divided by [math]n^{kn-1}[/math] (see 2°), and each term in the polynomial T is divided by [math]n^{k(n-1)}[/math], since both U and C are is divided by [math]n^k[/math]. 6°) As a result the numbers [math]U^n+C^n, A^n+B^n [=C^n], U^n[/math] are divided by [math]n^{kn-1+k(n-1)}[/math] [[math]=n^{2kn-k-1}[/math]]. 7°) Consequently, the number [math]U^n[/math] and [math]C^n[/math] are equal on the module [math]n^{2kn-k-2}[/math]. 8°) But then the number A+B finishes on [math]2kn-k-2 [>2kn-1][/math] zeros, that contradicts 3°. The case, when ABC not is divided by n, will be examined later. (On June 22, 2010)

FINAL (You will excuse me for the poor quality transfer) Proof of the FLT with the aid of Theorem about adjacent dividends: If the number [math]g=a^n+b^n[/math] (where a and b are natural) is divided by d, then the nearest to g numbers, which are divided by d, are the numbers [math](a-d)^n+b^n[/math] and [math](a+d)^n+b^n[/math]. (Simple proof of theorem is based on the lemma: “If the number [math]a^n+b^n[/math] is divided by prime d>1 and the numbers a, b and prime e, where [math]0<e<d[/math], are not divided by d, then the number [math](a+e)^n+b^n[/math] is not divided by d”, whose proof will be examined later.) Strictly proof of the FLT It follows from the equality 1°) [math]A^n+B^n=C^n[/math], where n>2 and [math]A, B, C (C>A>B)[/math] are natural, that 2°) [math]C/2<C/2^{1/n}<A<C[/math] (with A=B we have a equality: [math]A2^{1/n}=C^n[/math]), [math]B>1[/math] and 3°) [math]A+B<AB[/math]. Let us examine the inequality 4°) [math]F=D^n-(A^n+B^n)>0[/math], where D=A+B and, consequently, the number F is divided into AB. Let us find a number G adjacent with the number F and which is divided by AB. According to the theorem about adjacent dividends this there will be the number [math]G=(D-AB)^n-(A^n+B^n)[/math], which, accordingly 3°, is negative, and therefore between A+B and A+B-AB the entire value of the number D, satisfying conditions G=0 and G is divided into AB (with n>2) is absent. (With n=2 the negative number A+B-AB in terms of the absolute value it can prove to be less than the number A+B and the equality G=0 can be possible.) On June 6, 2010 Merged post follows: Consecutive posts mergedThus, for n>2 I proved inequality [math](A+B)^n-(A^n+B^n)>[0=]C^n-(A^n+B^n)[=0]>(A+B-AB)^n-(A^n+B^n)[/math], where all three numerical expressions are divided into AB. From the other side, according to the theorem about adjacent the divisible, extreme number-expressions as dividends on AB they are ADJACENT, i.e., between them there are no other dividends by AB! Here, strictly speaking, ALL proof of Fermat's last theorum. (Subsequently all attention it will be switched to the theorem about adjacent dividends.)

Half-Joke For relatively prime [math]A, B, C[/math] (where [math]C>A>B>1, AB[/math] is not divided by prime [math]m[/math]) and [math]m=n-1>1[/math] let us examine the equality 1°) [math]A^m+B^m-C^m=0[/math], or 2°) [math]A^m=C^m-B^m=(C-B)P, B^m=C^m-A^m=(C-A)Q[/math], from where 3°) [math]C^n-A^n-B^n= (P+Q)(C-B)(C-A)=D^n[/math]. To the proof of the FLT The number D is either 1) whole or 2) fractional. In the first case, as can easily be seen (taking into account that the number [math]C-B, C-A, P+Q[/math] are relatively prime), 4°) [math]C-B=x^{mn}, C-A=y^{mn}, P+Q=z^n[/math], and provability of the FLT is very real. Impression is created, that the case 2 also is demonstrated.

There is no proof. A study continues. For the amateurs of mathematics, managing general algebraic theory of FLT. Theorem 5 If in the prime base n>2 2-digit end of the s-digit number A is the 2-digit end of the n-th degree, not multiple n, then the number A has the equivalent form: [math]A=a^n-Pn^w[/math], where w are more than the given number v and the w-digit end of the number a^n is equal to A. Theorem 5a (generalized) If in the prime base n>2 t-digit end of the s-digit number A is the t-digit end of [math]n^{t-1}[/math]-th degree, not multiple n, then the number A has the equivalent form: [math]A=a^{t-1}-Pn^w[/math], where v are more than the given number and the w-digit end of the number [math]a^{t-1}[/math] is equal to A. The validity of these impressive theorems follows from the simplest fact: the k-th digit (from the end) in the number a and (k+1)-th digit in the number a^n (k>1) one-to-one are determined the values of each other (that it follows from the binomial of Newton and of fermat's litlle theorem in the prime base n>2). These two theorems make it possible to propose simple proof of FLT. Actually, from the equality 1°) [math]A^n+B^n=C^n[/math] (A, B, C are coprime and prime n>2) it follows that the 2-digit ends of the numbers A, B, C are the 2-digit ends of n-th degrees, not of multiple n (with ABC not multiple n), and therefore with [math]n^{w+1}>C^n[/math] have the equivalent form: 2°) [math]A=a^{n^2}-Pn^w, B=b^{n^2}-Qn^w, C=c^{n^2}-Rn^w[/math]. Let us substitute these values into 1° and, after the disclosure of Newton's binomials, on the [math]n^w[/math]-digit ends of the all numbers in the equality 1° we obtain the second equality: 3°) [math]a^{n^2}+b^{n^2}=c^{n^2}[/math]. Question: how many numbers in the numbers a, b, c?

Theorem about the dividers (Sorokin’s theorem) If integers A, B and A-B are relatively prime and not multiple prime n>2, and the t-digits end of the number R in the equality [math]A^n-B^n=(A-B)R[/math] in the base n is equal to 1, then the t-digits end of of each prime divider m of the number R is also equal to 1. Proof consists of several theorems. Merged post follows: Consecutive posts mergedI approach the publication of strict proofs of theorems. Theorem 1. If integers A, B and A-B are relatively prime and are not multiple by prime n>2, then the last digit of each prime divider m of the number R in the equality [math]A^n-B^n=(A-B)R[/math] is equal to 1 in the base n. Proof. It is known that under theorem conditions the number R is not divided by n and numbers A-B and R have no common divider. Let us show now that also the prime number q of the form of [math]q=p+1[/math], where p is not divided by n, is the divider of the number [math]A^n-B^n[/math] and is not the divider of the number R. According to Fermat's little theorem, number [math]A^p-B^p[/math] is divided by q. It is utilized the solution of the following linear diophantine equation: nx-py=1[/math]. The number [math]A^ {nx}-B^ {nx}[/math] is divided by m. But [math]A^{nx}-B^{nx}=A^{py+1}-B^{py+1}=(A^p)^y-(B^p)^y[/math], where the numbers [math]A^p[/math] and [math]B^p[/math] have (according to Fermat's little theorem) last digit equal to 1. Consequently, the number A-B is divided, but the number R is not divided by q. Thus, the numbers of form [math]q=p+1[/math], where p is not divided by n, are not the dividers of the number R. QED.

Block diagram of the last (dated March 16) project of the proof 1. On the last (different from zero) number of each prime divider of the number a we determine (with the aid of the Fermat's little theorem) the last digit of the number [math]a^(n-1)[/math] in the prime base n>2; this be 1. 2. Since in the equality [math]a^n+b^n=(a+b)R[/math] (and two analogous) the number R (not divided by n) is [math]r^n,[/math] then the two-digit end of the number R is 1 (or 01). 3. From this it follows (on the basis one of the S-theorems) that the two-digit ends of the numbers (not divided by n) [math]a^{n-1}, b^{n-1}, c^{n-1}[/math] are 1. 4. From this it follows (on the basis one of the S- theorems) that the two-digit ends of each of the simple dividers of the numbers (not divided by n) R and two others (not divided by n) is 1. 5. Since in the equality [math]a^n+b^n=(a+b)R[/math] (and two analogous) the number R (not divided by n) is r^n, then the three-digit end of the number R is 1 (or 001). Thus further - to infinity.

Theorem about transformation If in the prime base n the two-digit end of the number a (a is not divided by n) is a two-digit end of a certain number [math]a'^n[/math], then there is a number A^n with the end a, i.e., [math]a=A^n-Pn^k[/math], where [math]n^k>a[/math]. (Simple, but voluminous, the proof of Theorem is based on the fundamental theorem: In the numeration system in the prime base n>2 for any single-digit number a (a is not divided by n) and the given number d there is an only number g, that the last digit of the number ag is d. I do not have a possibility to refer to sources, but entire apparatus for the conversion of numbers was repeatedly presented by me on the sites of “mmonline” and “dxdy”. This entire the simple apparatus, repeatedly checked by participants in the forums, I will present again, but after the completion of proof and of his acknowledgement by the true under the assumption to faithfulness to my theory of numeration in the base with the prime base.) Corollary from Theorem about transformation If in the prime base n the two-digit end of the number a (a is not divided by n) is a two-digit end of a certain number [math]A^n[/math], then there is this number [math]a^{n^t}[/math], that [math]a'^{n^t}=a+Pn^k[/math], where k is as desired great. (Proof is based on the simple lemma: Last k digits of the number a do not influence value (k+1)-th digit of the number a^n.) Proof of the FLT Let us represent the number a, b, c (thhey are not divided by n) in the equality 1°) [math]a^n+b^n=c^n[/math], according to corollary about the Theorem about transformation: 2°) [math]a=a'^{n^t}-Pn^k, b=b'^{n^t}-Qn^k, c=c'^{n^t}-Rn^k[/math], where k is as much as desired great. Then as much as desired the long ends of the numbers [math]a^{n-1}, b^{n-1}, c^{n-1}[/math] (in the equality of 1°) finish to 1 and, therefore, the number of [math]u=a+b-c[/math] finishes to as much as desired number of zeros (if abc not multiply n and if one of the numbers a, b, c is not divided by n). It is understandable, the text of even damp and on the course of events it will be refined

Proof of Fermat's last theorem Thus, Let us assume that for relatively prime natural A, B, C (for the certainty A>B and AB(A-B) is not divided by n) and prime n>2 1°) [math]A^n+B^n=C^n[/math], where, as is known, 2°) [math]A^n-B^n=(A-B)R[/math]; 3°) the number C-B, C-A, A-B and R are relatively prime; 4°) [math]C-B=a^n, C-A=b^n[/math]; from where 5°) [math]a^n-b^n=A-B[/math]; 6°) each prime divider m of the number R takes the form m=pn+1. Proof of the FLT Let us take any prime divider m of the number R. Then, according to Fermat's little theorem, the numbers 7°) [math] (C-B)^{pn}-(C-A)^{pn}[/math] and [math]a^{pn}-b^{pn}[/math], or [math] (C-B)^{pn-1}-(C-A)^{pn-1}[/math], are divided by m. But so the numbers in the pairs (C-B, C-A) and (pn, pn-1) are relatively prime, then, according to the Lemma (its proof three times was brought on the forum of dxdy), the number m is the divider of the number A-B, that contradicts to 3°. FLT is proven. *** (The proof of the Lemma, based on the formula of the solution of linear diophantine equations, will be given after the elimination of all other questions, which arose in the proof of the FLT.) Merged post follows: Consecutive posts mergedProof of Fermat's last theorem (the completing text) Let us assume that for relatively prime integers A, B, C (for the certainty A>B and AB(A-B) is not divided by n) and prime n>2 1°) [math]A^n+B^n=C^n[/math], where, as is known, 2°) [math]A^n-B^n=(A-B)R[/math]; 3°) the number C-B, C-A, C, A-B and R are relatively prime; 4°) [math]C-B=a^n, C-A=b^n[/math]; from where 5°) [math]a^n-b^n=A-B[/math]; 6°) each prime divider m of the number R takes the form m=pn+1. Proof of the FLT Let us take any prime divider m of the number R. Then the numbers [math]A^n-B^n[/math] and therefore 7°) [math]A^{n(p-1)}-B^{n(p-1)} [/math] and, according to Fermat's little theorem, 8°) [math]a^{pn}-b^{pn}[/math], или [math](C-B)^{n(p-1)}-(C-A)^{n(p-1)}[/math] 9°) are divided by m. But so the number C is relatively prime with the number (A-B)R, then, according to lemma (see the Appendix), the numbers 7° and 8° are relatively prime, that contradicts 9°. The case, when CB(C+B) is not divided by n, proves analogously. *** Appendix Lemma If the number C is relatively prime with the number [math]A^n-B^n=(A-B)R[/math], where n>2, then the numbers [math]A^n-B^n[/math] and [math](C-A)^n-(C-B)^n[/math] are relatively prime relative to the dividers of the number R. Unfortunately, the author does not have available the proof of Lemma. However, it is possible that P.Fermat found its proof in Diophante's “Arithmetic”. A study strictly of great theorem (without considering Lemma) ceases on this. Merged post follows: Consecutive posts mergedAnalogous proof, but WITHOUT the Lemma. Let us assume that for relatively prime integers A, B, C (for the certainty A>B and AB is not divided by n) and prime n>2 1°) [math]A^n+B^n=C^n[/math], where, as is known, 2°) the number C-B and C-A are relatively prime and [math]C-B=a^n, C-A=b^n[/math]; from where 3°) [math]a^n+b^n=(a+b)R[/math]; 4°) the number R contains the prime divider m of the form [math]m=pn+1[/math] (author's proof uses a formula of the solution of linear diophantine equation). Proof of the FLT Let us take any prime divider m of the number R. Then the numbers [math]a^n+b^n[/math], or [math](C-B)+(C-A) [/math], and therefore 5°) [math]D=(C-B)^{n(p-1)}+(C-A)^{n(p-1)}[/math], where [math]n(p-1)[/math] is odd (!), and, according to Fermat's little theorem, 6°) [math]E=(C-B)^{n(p-1)}-(C-A)^{n(p-1)}[/math], 7°) are divided by m. But so the numbers (C-B) and (C-A) are relatively prime (without considering dividers 2), then the numbers D and E have no common factors, which contradicts 7°. The case, when CB is not divided by n, proves analogously. FLT is proven. On March 12, 2010 =========================== END

I advance two similar S-hypotheses (“Sorokin’s hypothesis”). S-hypothesis-1 (confirmed by concrete calculations on the computer): In the equality [math]A^{n^t}+B^{n^t}=(A^{n^{t-1}}+B^{n^{t-1}})R[/math], where - prime n> 2, - integers A and B are relatively prime, (t-1)-digit end of each prime divider of the number R (excluding n) in the base n is equal to 1. For example (for n=3): [math]8^3-1=(8-1)*73=(8-1)*(8*9+1) [/math]; [math]8^3+1=(8+1)*57=(8+1)*(3*19)=(8+1)*[3*(2*9+1][/math]; and so forth. S-hypothesis-2 (more questionable): In the equality [math]A^n+B^n=(A+B)R[/math], where - prime n> 2, - integers A and B are relatively prime, - their k-digit ends are the k- digit ends of some numbers of a^n and b^n, - the k-digit end of the number R is equal to 1, the k-digit end of each prime divider of the number R (excluding n) in the base n is equal to 1.

Clean proof of the FLT. A special case. Let for relatively prime A, B, C (for the certainty [math]AB(A-B)[/math] is not divided by n) and prime [math]n>2[/math] 1°) [math]A^n+B^n=C^n[/math]. Are known the following properties of the Fermat’s equality: 2°) [math]C-B=a^n=p[/math]; [math]C-A=b^n=q[/math]; 3°) [math]p^n-q^n=(p-q)R[/math], where each prime base m of the number R takes the form [math]m=kn+1[/math]. Let us prove the FLT for k not divided by n. 4°) According to Fermat's little theorem, the number [math]a^{kn}-b^{kn}[/math], or [math]p^k-q^k[/math], or [math] (p-q)T[/math] (just as the number [math]p^n-q^n=(p-q)R[/math]) contains divider m. But, as it is known (and that easy to show with the aid of the linear diophantine equation [math]kx-ny=1[/math]), if p and q relatively prime, p-q is not divided by n, k and n are relatively prime, then the numbers T and R are relatively prime, that contradicts 4°. *** Thus, for the completion of proof of the FLT it remains to show that the case, when k is divided by n, is also contradictory. Merged post follows: Consecutive posts mergedWith probability 99% under the conditions of theorem, the number k is divided by n. If this hypothesis is true (it is possible that it is proven), then the first case has no benefit. Merged post follows: Consecutive posts mergedIt is very probable that as the basis of its proof of the great theorem P.Ferma it placed precisely this theorem: For mutually-prime numbers a and b each prime divider m (with exception n) of the number [math]R=(a^{n^{t+1}}+b^{n^{t+1}})/(a^{n^t}+b^{n^t})[/math] has the form of [math]m=pn^{t+1}+1[/math]. For example: [math](2^9-1)/(2^3-1)=73=8[/math]* [math]9+1[/math]. It is interesting, is there a simple proof of this theorem? Apparently, P.Ferma proved also the stronger theorem: If the t-digit ends of the numbers A and B are the t-digit ends of some numbers a^n and b^n, then the t-digit end of EACH prime divider R in the equality [math]A^n+B^n=(A+B)R[/math] is equal to 1. For example: [math]=(***9+***8)(9^2-72+8^2)=73=***8[/math]*[math]9+1[/math]. If this theorem is true, then FLT proves from the force into ten lines: If the number A+B-C is divided by [math]n^t[/math], then it “automatically” is divided also into [math]n^{t+1} [/math]. And so to infinity. But the very important: two the theorems indicated generate large NEW region in the theory of the numbers.

Proof of the FLT (project) Let us assume that for relatively prime integers [math]A, B, C[/math] and prime n>2 1°) [math]A^n+B^n=C^n[/math], from here 2°) [math]A^n-B^n=C^n-2B^n=D[/math]. 3°) It’s known that the number [math]R[/math] in the equality [math]A^n-B^n=(A-B)R[/math] has the prime divider [math]m[/math] of the form [math]m=pn+1[/math]. Let us show, however, that the number [math]D=C^n-2B^n[/math] is not divided by m. According to Fermat's little theorem, the number 4°) [math]E=C^{pn}-2^{pn}B^{pn}[/math] (as the number D) is divided by prime m. However, the numbers D and E are relatively prime relative to m. Let us show this. ……………….. (Proof based on the formula of linear diophantine equation.)

It is seemed that at the basis of my last proof the lemma must lie: For any rational h and assigned entire (A+B), where A is not equal to B, there is such t, that [math](A+t)^n+(B-t)^n=(A+B)R=h[/math]. And it is now necessary to show that with the given three integers (C-B), (C-A), (A+B) there are such rational P, Q, R (at least in the form common fractions), that [math](C-B)P+(C-A)Q=n^nR[/math] [or [math] (C-B)P+(C-A)Q=nR[/math]], where (A, B, C) is a solution of Fermat’s equation. OR: And it is now necessary to show that with the given three integers (c-b), (c-a), (a+b) there are such rational P, Q, R (at least in the form common fractions), that [math](c-b)P+(c-a)Q=n^nR[/math] [or [math](c-b)P+(c-a)Q=nR[/math]], where c-b=C-B, c-a=C-A and (A, B, C) is a solution of Fermat’s equation. Merged post follows: Consecutive posts mergedInteresting moment in light of the last idea – equality [math](C-B)P-(C-A)Q-(A-B)T=0, or A^n=B^n+D^n[/math] [from here [math]D^n=A^n-B^n=(A-B)S, U'=B+D-A[/math]], where [math]A^n+B^n-C^n=0[/math] is Fermat’s equality, the numbers A, B, C are whole end the number D is fractional.

It is seemed that at the basis of my last proof the lemma must lie: For any rational h and assigned entire (A+B), where A is not equal to B, there is such t, that [math](A+t)^n+(B-t)^n=(A+B)R=h[/math]. And it is now necessary to show that with the given three integers (C-B), (C-A), (A+B) there are such rational P, Q, R (at least in the form common fractions), that [math](C-B)P+(C-A)Q=(A+B)R[/math].

Very simple idea Proof of the FLT (yet not acknowledged with mathematical association) Let us assume that for relatively primes [math]A, B, C (C>A>B>1)[/math] the equality 1°) [math]A^n+B^n=C^n[/math] exists. Let the number AB not is divided by n. 2°) It’s known that if [math]C=C'n^{kn}[/math], where C' not is divided by n, then [math]A+B=C"n^{kn-1}[/math], where C" not is divided by n. Let us compose the new equality 3°) [math]a^n+b^n=c^n[/math] with the condition: 4°) [math]c-a=C-A, c-b=C-B, a+b=n^n[/math] – if A+B of odd, and [math]a+b=(2n)^n[/math] – if A+B of even. In the case of odd A+B, from these relationships we find: [math] (c-a)+(c-b)+(a+b)=2c=(C-A)+(C-B)+n^n[/math]; [math] (c-a)-(c-b)+(a+b)=2b=(C-A)-(C-B)+n^n[/math]; [math]-(c-a)+(c-b)+(a+b)=2a=-(C-A)+(C-B)+n^n[/math], where right sides are even and, therefore, the numbers a, b, c are WHOLES. From the other side, accordingly 2°, solution [math] (a, b, c)[/math] of the equation 3° are not INTEGRAL. In the case of even A+B the conclusion is analogous. And we arrived to the contradiction.

Today I left to the theorem, whose proof indicates the gap of the closure of logic of FLT and, therefore, the inaccuracy of theorem (if the same exists) about the absence of elementary proof of FLT. Here is the formulation of theorem. Theorem For the relatively prime numbers a, b, c, d, greater than 1, the numbers [math]p=(a^n-b^n)/(a-b)[/math] and [math]q=(c^{n-1}-d^{n -1})/(c-d)[/math], where n is simple, sont relatively prime. Theorem proves with the aid of the following lemma (and lemma itself it proves with the aid of the linear diophantine equations): Lemma For the relatively prime numbers a and b everything - with exception perhaps of only (namely the number n) - the prime dividers m of the number [math]p=(a^n-b^n)/(a-b)[/math] have the form of m=tn+1. Merged post follows: Consecutive posts mergedFollowing lemma “Numbers [math]A+B-C (=U> 0)[/math] and [math]A-B[/math] under the conditions of the FLT are relatively prime”. (I will resemble: the numbers [math]A, B, C, C-B, C-A, A-B, C+B, C+A[/math] are relatively prime and [math]A^n+B^n=C^n[/math].) I proved thus: Let us assume that the first number contains divider d of the number A-B. Then both numbers in the following pairs are divided by d: [math]2A-C [=A+B-C+(A-B)][/math] and [math]A-B[/math]; [math](2A)^n-C^n [=(2A-C)R][/math] and [math]2A^n-C^n [=A^n-B^n=(A-B)T][/math]; [math](2^n-2)A^n [=(2A)^n-C^n-(2A^n-C^n)][/math] and [math]2A^n-C^n[/math]; [math]2(2^{n-1}-1)[/math] and [math]2A^n-C^n[/math] [since [math]A^n[/math] and [math]A^n-B^n[/math] relatively prime]. It is evident from this that the common divisors of the numbers A+B-C and A-B can be located only among the dividers of number [math]2(2^{n-1}-1)[/math]. *** Strictly speaking, proof of the FLT in essence is already published: this is the Lemma given above. If we investigate the passages from the pair to the pair (in these passages THE NEW prime dividers from If we investigate the passages from the pair to the pair (in these passages THE NEW prime dividers from the set of dividers of the number [math](2^{n-1}-1)[/math] CANNOT appear in principle!!!), then it is possible to note that ALL dividers of the number [math](2^{n -1}-1)[/math] - including n!!! - are dividers of the number A-B. And now either pair of relatively prime numbers A+B (multiple n) and A B, or all three relatively prime numbers A-B, C+B, C+A are divided by n. In my opinion "arrived"! I prepare the complete text of proof of the FLT.

Proof of the Fermat’s Last Theorem. Project for the reflection Let us assume that for the relatively prime integers A, B, C (C>A>B>0) and prime n>2 1°) [math]A^n+B^n-C^n=0[/math], where, as can easily be seen, the number [math]A^n+B^n, A^n-B^n, C^n-B^n, C^n+B^n, C^n-A^n, C^n+A^n[/math] are relatively prime. For the proof is used simple lemma about infinity of the set of primee numbers m of the form m=2kn+1 (proof it descends). Let us thus, take prime [math]m=2kn+1>2C^n[/math]. Then, according to fermat's small theorem, the number [math]A^{2kn}-B^{2kn}, C^{2kn}-B^{2kn} n, C^{2kn}-A^{2kn}[/math] in the base m finish with 0, from what it follows that one of the algebraic cofactors in each of these three numbers finish with 0 (i.e. they are multiple m). Let us assume that its are three following cofactors: 2°) [math]C^{kn}-B^{kn} [=(C^n-B^n)P], C^{kn}-A^{kn} [=(C^n-A^n)Q], A^{kn}-B^{kn} [=(A^n-B^n)R][/math]. I advance the following hypothesis: If the numbers c-a and c-b relatively prime and are more than 1, then the number [math]p=(c^n-a^n)/(c-a)[/math] and [math]q=(c^n-b^n)/(c-b)[/math] also relatively prime. If hypothesis is confirmed, then at least one of the numbers [math]C^n-B^n, C^n-A^n, A^n-B^n[/math] is multiple m, that is impossible, since [math]m>2C^n[/math]. The matter remained for “the small”…