Victor Sorokine

Proof of Fermat's last theorem (Final text) Basic properties (known from 17 centuries) of the Fermat’s equality: 1°) [math]A^n+B^n=C^n[/math] for the base case: 2°) prime n>2 (in this proof [math]n>4[/math]), 3°) integers A, B, C are relatively prime; 4°) a, b, c are greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B ); 5°) p, q, r are second cofactors in the numbers A, B, C: [math]A=ap[/math], [math]B=bq[/math], [math]C=cr[/math], where 6°) the number in the pairs (a, p), (b, q), (c, r) relatively prime; 7°) the number [math]U=A+B-C[/math] is divisible into abc; 8°) if [math]ABC[/math] is not divisible into n (case 1), then [math]C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B=c^n[/math]; 9°) if - for example, C - is divided into n (case 2), [math]then C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B= (c^n): n[/math]. 10°) is not difficult to see (taking into account 5°-8°) that the number U does not have dividers of the number pqr [that evidently from the equalities [math]U=ap-(C-B )=ap-a^n[/math], [math]U=bq-(C-A)=bq-b^n[/math], [math]U=(A+B )-cr=c^n-cr)[/math]]. *** Proof of the FLT in the numeration system with the prime base n>2 According to the theorem about the dividers of exponential trinomial, the set of prime dividers (or prime bases) in the number [math]U=A+B-C[/math] does not exceed the limits of the set of prime dividers in the number [math]nABC[/math]. In this case, as can be seen from 4°-5°, numbers a, b, c are the dividers of the number U, in this case the number U is not divisible into the number of abcd, where d is any divider of the number of abc. Furthermore, the prime dividers of the number [math]ABC[/math], which are not the dividers of the number nabc, are not the dividers of the number U, therefore, they are not dividers of the number [math]nABC[/math]. As a result under the conditions of the FLT we have a equality: 11°) [[math]U=[/math]] [math]A+B-C=abcn^t[/math]. However, after term-by-term multiplication equality 1° by [math]2^{nn}[/math] equality 11° (with n>4) is disrupted, since the numbers [math]C-B[/math] [[math]=a^n[/math]], [math]C-A[/math] [[math]=b^n[/math]], [math]A+B[/math] [[math]=c^n[/math]] and, therefore, the left side of the number U are multiplied by [math]2^n[/math], but the numbers a, b, c are multiplied altogether only by 2. As a result the right side of the equality 11° is multiplied by 8. But if, for example, the number C is divisible into n, then in the formula [math]c^n=(A+B ) [/math]n the number [math]A+B[/math] is multiplied by [math]2^n[/math]: 12°) [math]c’^n=[(A+B)2^n]n=[(A+B)n]2^n=[c^n]2^n=[2c]^n[/math], which again leads to the disturbance of the equality of 11°, and truth of the FLT is present. (“Difficult” case itself: n=3 is degenerate and proves separately.)

Estimation of the situation I consider with the large confidence that I proved the following assertions: 1. Theorem about the greatest common divisor of exponential binomials, 2. Theorem about the dividers of exponential trinomial, 3. Formula for the Fermat’s equality: A+B-C=abcn^k, 4. FLT for the case: prime n> 4 and ABC not is divided by n. The search for the perfect proofs of the remaining cases continues. +++ For Matthew: I will try to consider your observations. The many thanks! Linguistic proof-reading Theorem about the dividers of the exponential trinomial Theorem If integers A, B, C are relatively prime and the number [math]d=A^n+B^n-C^n[/math] is divisible, and the number of [math]nABC[/math] is not divisible by a prime number q, then the number [math]U=A+B-C[/math] is not divisible by q. Proof by the method from the contrary (i.e. let us assume that the number U is divisible into q) 1°). Let us examine the numbers A, B, C, D, U in the base-q numeral system and let us multiply the number D by such a number [math]g^n[/math] that the number [math]gB[/math] finishes by digit 1 (see A1°). 2°). Let us write down the new values of the numbers A, B, C in the form: [math]A=qx+a[/math], [math]B=qy+1[/math], [math]C=qz+c[/math] (where a, b, c are the last digits in the numbers A, B, C and, on the assumption, the number [math]u=a+1-c[/math] is divisible by q) and let us substitute these values into the formula for D. 3°) Since the number D is divisible by q, then, therefore, the number [math]d=a^n+1-c^n[/math] is divisible by q. 4°) Since the number U is divisible by q, then, therefore, the number [math]e=a^q+1-c^q[/math] is divisible by q. 5°) Consequently, the number [math]d-e=(a^n-c^n)-(a^q-c^q)[/math] also is divisible by q. 6°). However (according to the Theorem about the greatest common divisor of two exponential binomials – see A6°) the greatest common divisor of the two numbers [math] (a^n-c^n)[/math] and [math](a^q-c^q)[/math] is the number [math]a-c[/math], which, as can be seen from 2°, is not divisible by q. Thus the Theorem is proven. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Appendix A1° - See the multiplication tables of two numbers written in a numeral system with a prime base. A6° - Lemma (theorem about the greatest common divisor of exponential binomials). +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Theorem about the common divisor of the exponential binomials Lemma (theorem about the greatest common divisor of exponential binomials) If, given two relatively prime integers a and b and relatively primes n (> 2) and q (> 2) both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] are divisible by a prime number q, then the number [math]a-b[/math] is divisible by q. Proof by the method from the contrary (i.e. let us assume that the number [math]a-b[/math] it is not divisible by q, but both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] are divisible by q) Let us examine two numbers D and E, which, obviously, are divisible by q: [math]D=a^{xn}-b^{xn}[/math] and [math]E=a^{yq}-b^{yq}[/math], where x and y are the solution of the linear diophantine equation [math]xn-yq=1[/math], or [math]D=a^d-b^d[/math] and [math]E=a^e-b^e[/math], where [math]d-e=1[/math]. From the multiplication [math]E(a+b )=a^{e+1}-b^{e+1}+ab(a^{e-1}-b^{e-1})=D+abD_2[/math] we can see that the number [math]D_2=a^{e-1}-b^{e-1}[/math] is also divisible by q. And so on: From the multiplication [math](E_2)(a+b )=a^{e-1}-b^{e-1}+ab(a^{e-3}-b^{e-3})=D_2+abD_3[/math] we can see that the number [math]a^{e-3}-b^{e-3}=D_3[/math] is also divisible by q. And so on, up to the number [math] (a^2-b^2)(a+b )=a^3-b^3+ab(a-b )[/math], where the number a-b[/math] is divisible by q, QED. *** This lemma appears to be the key in the proof of the Theorem about the dividers of exponential trinomial. So far I can not find any evidence that the Lemma and the Theorem (two fundamental theorems in the theory of the numbers) were proven earlier.

Popular science communication about proof of the FLT After the countless attempts to find contradiction between THE EXISTING elements of the Fermat’s equality in me gradually began to ripen idea about the fact that the contradiction should be looked between the elements, WHICH ARE ABSENT in the equality farm. This idea finally led to the study “nonexistent” prime dividers of the number [math]U=A+B-C[/math]. And actually, soon began to appear the signs of the fact that, besides the dividers of the numbers A, B, C yes degree itself n, no other dividers the number U contains! Without being limited to itself Fermat’s equality, I began to search for this property in any trinomials and find his proof for all exponential trinomials [math]D=A^n+B^n-C^n[/math] with relatively prime A, B, C. It is shorter, the number U in the trinomial D does not exceed the number [math]ABCn^k[/math], but in the Fermat’s equality is that less: [math]U=abcn^k[/math], where a, b, c is the greatest common divisors in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B ). However, most interesting proved to be very equality [math]A+B-C=abcn^k[/math], in which k>0 and last significant digits in the left and right sides of the equality, where they are obligated to be equal. BUT!!! During the multiplication of the Fermat’s equality by [math]2^n[/math] last significant digits in the numbers A, B, C and U are multiplied by 2, while the last significant digits in the number [math]abcn^k[/math] is multiplied by 8 (if among the numbers A, B, C there is no multiple n) or, at the worst, by 4 (if among the numbers A, B, C one multiply n), which leads to THE INEQUALITY of the last significant digits in “the equality” [math]A+B-C=abcn^k[/math]! Let us note that no elliptical functions for the proof of this contradiction it was required! I do not expect that the proof of the FLT will be in the very near future acknowledged by mathematical leaders (“This it cannot be, because this it cannot be ever!!! ”). But to the beginning mathematicians there is above how to think… I will regret, which interfere withs poor transfer clear understanding. I hope that this problem will be soon solved.

Proof of Fermat's last theorem Basic properties (known from 17 centuries) of the Fermats equality: 1°) [math]A^n+B^n=C^n[/math] for the base case: 2°) prime n> 2, 3°) integers A, B, C are relatively prime; 4°) a, b, c are greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B ); 5°) p, q, r are second cofactors in the numbers A, B, C: [math]A=ap[/math], [math]B=bq[/math], [math]C=cr[/math], where 6°) the number in the pairs (a, p), (b, q), (c, r) relatively prime; 7°) last digits in the numbers p, q, r are 1; 8°) the number [math]U=A+B-C[/math] is divided into abc; 9°) if [math]ABC[/math] is not divided into n (case 1), then [math]C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B=c^n[/math]; 10°) if - for example, C - is divided into n (case 2), then [math]C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B= (c^n): n[/math]. 11°) is not difficult to see (taking into account 5°-8°) that the number U does not have dividers of the number pqr [that evidently from the equalities [math]U=ap-(C-B )=ap-a^n[/math], [math]U=bq-(C-A)=bq-b^n[/math], [math]U=(A+B )-cr=c^n-cr)[/math]]. Set of instruments: 12°) The Theorem about the dividers of exponential trinomial. *** Proof of the FLT in the numeration system with the prime base n>2 According to the theorem about the dividers of exponential trinomial, the set of prime dividers (or prime bases) in the number [math]U=A+B-C[/math] does not exceed the limits of the set of prime dividers in the number [math]nABC[/math]. In this case, as can be seen from 4°-5°, numbers a, b, c are the dividers of the number U, in this case the number U is not divided into the number of abcd, where d is any divider of the number of abc. Furthermore, the prime dividers of the number [math]ABC[/math], which are not the dividers of the number abc, are not the dividers of the number U, therefore, are not dividers of the numbers of [math]nABC[/math]. All these dividers, as is known, finish to digit 1. Thus, we have a equality: [math]U=A+B-C=un^k=abcn^t[/math], in which the last significant digits of the numbers un^k and abcn^t are equal. However, this equality is disrupted after the multiplication of Fermats equality by [math]2^n[/math]. Actually, during the multiplication by [math]2^n[/math] the last significant digits of the number [math]un^k[/math] is multiplied by 2, and the last significant digits of the number [math]abcn^t[/math] is multiplied - by 8 in the case of 1 (if ABC is not divided into n), - by 4 in the case of 2 (if ABC is divided into n), since the last significant digits of each of the numbers a, b, c are multiplied by 2. And we arrived at the contradiction.

Thank! A - Appendix A1°. See the multiplication tables of two numbers in the numeration system with the prime base. A6°. Lemma (theorem about the greatest common divisor of exponential binomials). ++++++++++++++++++++++++ Theorem about the common divisor of the exponential binomials Lemma (theorem about the greatest common divisor of exponential binomials) If with relatively prime integers a and b and relatively primes n (> 2) and q (> 2) both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] are divided into the prime number q, then the number [math]a-b[/math] is divided into q. Proof by the method from the contrary (i.e. let us assume that the number a-b[/math] it is not divided into q, but both numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] divided into q) Let us examine two numbers D and E, which, obviously, are divided into q: [math]D=a^{xn}-b^{xn}[/math] and [math]E=a^{yq}-b^{yq}[/math], where x and y are the solution of linear diophantine the equations [math]xn-yq=1[/math], or [math]D=a^d-b^d[/math] and [math]E=a^e-b^e[/math], where [math]d-e=1[/math]. From the multiplication [math]E(a+b )=a^{e+1}-b^{e+1}+ab(a^{e-1}-b^{e-1})=D+abD_2[/math] is evident that the number [math]D_2=a^{e-1}-b^{e-1}[/math] also is divided into q. It is further analogous: From the multiplication [math](E_2)(a+b )=a^{e-1}-b^{e-1}+ab(a^{e-3}-b^{e-3})=D_2+abD_3[/math] is evident that the number [math]a^{e-3}-b^{e-3}=D_3[/math] also is divided into q. And so on to the number [math](a^2-b^2)(a+b )=a^3-b^3+ab(a-b )[/math], where the number a-b is divided into q, QED. *** This lemma appears the key in the proof of the Theorem about the dividers of exponential trinomial. Thus far there is no evidence of the fact that the Lemma and Theorem (two fundamental theorems in the theory of the numbers) were proven earlier.

As to say correctly in English: Number 6 is divided into prime number 3? Number 6 is divided by prime number 3? Number 3 is a divider of number 6?

On the contrary, prime number q is divider of three numbers: [math]A^n+B^n-C^n[/math], [math]A^q+B^q-C^q[/math], and [math]A+B-C[/math]. +++++++++++ Theorem about the dividers of the exponential trinomial Theorem If integers A, B, C are relatively prime and the number [math]d=A^n+B^n-C^n[/math] is divided, and the number of [math]nABC[/math] is not divided into the prime number q, then the number [math]U=A+B-C[/math] is not divided into q. Proof by the method from the contrary (i.e. let us assume that the number U is divided into q) 1°). Let us examine the numbers A, B, C, D, U in the numeration system on the base q and will multiply the number D by this number [math]g^n[/math], that the number [math]gB[/math] finishes by digit 1 (see A1°). 2°). Let us write down the new values of the numbers A, B, C in the form: [math]A=qx+a[/math], [math]B=qy+1[/math], [math]C=qz+c[/math] (where a, b, c are last digits in the numbers A, B, C and, on the assumption, the number [math]u=a+1-c[/math] is divided into q) and let us substitute these values into the formula for D. 3°) Since the number D is divided into q, then, therefore, the number [math]d=a^n+1-c^n[/math] is divided into q. 4°) Since the number U is divided into q, then, therefore, the number [math]e=a^q+1-c^q[/math] is divided into q. 5°) Consequently, the number [math]d-e=(a^n-c^n)-(a^q-c^q)[/math] also is divided into q. 6°). However (according to the Theorem about the greatest common divisor of two exponential binomials see A6°) the greatest common divisor of the two numbers ([math]a^n-c^n[/math]) and ([math]a^q-c^q[/math]) is the number a-c, which, as can be seen from 2°, is not divided into q. Thus Theorem is proven.

Theorem about the dividers of exponential trinomial (precise formulation) If natural A, B, C – relatively prime and the number [math]A^n+B^n-C^n[/math] is divided, but the number of [math]nABC[/math] is not divided into the prime number q, then the number [math]U=A+B-C[/math] is not divided into q.

I yet did not understand the reason for contradiction with dividers 31 and 47. Searches continue. There are hypotheses.

Thanks! This IS VERY interesting! I will in greater detail answer by evening.

You are completely right, and your counterexample is excellent! Enormous thanks!!! I understood the reason for marriage. True theorem is formulated as follows: The number [math]A^n+B^n-C^n[/math], where A+B-C is divided into prime n, and A, B, C - relatively prime, and A+B-C and ABC are not equal to zero, it does not contain prime bases beyond the limits of their set in the number ABC. Difference as if is small, but is fundamental.

Actually, I proved the following theorem: The number [math]A^n+B^n+C^n[/math], where A+B+C is divided into prime n, and A, B, C - relatively prime, and A+B+C and ABC are not equal to zero, it does not contain prime bases beyond the limits of their set in the number ABC.

Materials to the proof of the formula [math]A+B-C=abc[/math]. T1. Lemma (fundamental theorem, main tool for proof of the FLT). With the relatively prime numbers a and b and relatively prime n (>2) and q (>2) the greatest common divisor of the numbers [math]a^n-b^n[/math] and [math]a^q-b^q[/math] is [math]a-b[/math]. Proof. Let us examine the numbers [math]D=a^{xn}-b^{xn}[/math] and [math]E=a^{yq}-b^{yq}[/math], where [math]xn-yq=1[/math] (see linear they are diophantine the equation), or [math]D=a^d-b^d[/math] and [math]E=a^e-b^e[/math], where [math]d-e=1[/math]. From the multiplication [math]E(a+b )=a^{e+1}-b^{e+1}+ab(a^{e-1}-b^{e-1})=D+abD_2[/math] is evident that the number [math]a^{e-1}-b^{e-1}=D_2[/math] also is divided into q. Is further analogous: From the multiplication [math] (E_2)(a+b )=a^{e-1}-b^{e-1}+ab(a^{e-3}-b^{e-3})=D_2+abD_3[/math] is evident that the number [math]a^{e-3}-b^{e-3}=D_3[/math] also is divided into q. And so on to the number [math](a^2-b^2)(a+b )=a^3-b^3+ab(a-b )[/math], where the number a-b is divided into q, QED.

Returning to the prime dividers q, different from the dividers of the number abcn. 0°. Earlier was given simple and brief proof of the FLT (for n>2) for the case [math]A+B-C=abc[/math], where a, b, c - greatest common divisors respectively in the pairs of numbers (A, C-B ), (B, C-A), (C, A+B ), where A, B, C - relatively prime numbers of the Fermat’s equality. Now let us prove (in the theses) the equality [math]A+B-C=abc[/math], i.e., let us show that if [math]U=A+B-C[/math] is divided into prime n and prime q (which is not been the divider of the number [math]ABC[/math]), then the number [math]A^n+B^n-C^n[/math] is not divided into q. (Obvious that the number [math]U=A+B-C[/math] in theFermat’s equality does not have prime dividers from their set in the number [math]ABC[/math], which are not been the dividers of the number abc.) 1°. First of all, with the aid of the multiplication of the number U by the appropriate number g we convert the last digit of the number B, recorded in the numeration system on the base q, into 1. (possibility of this conversion it was repeatedly shown by the author on the mathematical forums.) 2°. If the sum of the remainders [math]A'+1-C'[/math] from the division of the new values of the numbers A, B, C on q (or last digits in the numbers A, B, C in the numeration system on the base q) is equal to 0, then let us replace positive number [math]A'[/math] by the negative number [math]q-A'[/math]: [math]1-a-c=q[/math], where a=[math]q-A'[/math], of [math]c=C'[/math],.. 3°. …in this case the numbers A, B, C can be recorded in the form: [math]A=qx-a[/math], [math]B=qy+1[/math], [math]C=qz-c[/math]. 4°. Let us substitute the values of the numbers A, B, C from 3 into the Fermat’s equality, from where, after the disclosure of Newton's binomials, we find that the number [math]d=1-a^n-c^n[/math] is divided into q. 5°. Known also that if [math]1-a-c=q[/math] (q is prime), then the number [math]e=1-a^q-c^q[/math] is divided into q. 6°. But then the difference in [math]d-e=(a^q-c^q)-(a^n-c^n)[/math] also is divided into q. 7°. However, it is known (which was repeatedly shown the author on the forums), that the greatest common divisor of the numbers [math](a^q-c^q)[/math] and [math](a^n-c^n)[/math] is the number [math]a-c[/math], which, as can be seen from 2, into q is not divided. Thus, taking into account 0°, the solution of Fermat’s equation in the integers does not exist. FLT is proven. ================= The author is ready to answer any questions, which are concerned this proof.

Post is rejected because of the inaccuracy.

The third it is not given! According to the basic law of the Logic, the given above lemma A) or is true, B ) or is not true. In the case A) the number [math]a^n+b^n-c^n[/math] (=0!) is not divided completely by an integer q. Therefore, in this case FLT is true. In the case B ) the finite number U ([math]=a+b-c[/math]) is divided completely by ALL integers q. Therefore, FLT is true in this case. But THE THIRD IT IS NOT GIVEN!!!

Simplification of the proof Base lemma At the basis of elementary proof of the FLT lies the following simple lemma: Lemma If the number a, b, c relatively prime and the number U=a+b-c is divided by prime n>2, the number [math]a^n+b^n-c^n[/math] is not divided by prime q, different from the dividers of the number abcn. Proof of Fermat's last theorem Possibly Fermat's equality contradicts to lemma.

Repetition of that passed At the basis of proof of the FLT lies simple lemma 1 (according to all signs, previously not known): If [math]a^n+b^n-c^n=abcd[/math], where prime n>2 and the numbers a, b, c are relatively prime, the number d=1. At the basis of proof the simple lemma lies: If [math]a+b-c[/math] is equal to prime e (e not equally to n>2), the number [math]a^n+b^n-c^n[/math] is not divided by e. With the aid of the lemma 1 proof of the FLT in the case of 1 (ABC not divided by prime n>2) it is exceptionally simple: The equality [math]A^n+B^n=C^n[/math] is contradictory, since the contradictorily second equality [math]A+B-C=abc[/math], where the left side is divided by n, but right is not divided. (Here [math]C-B=a^n[/math], [math]C-A=b^n[/math], [math]A+B=c^n[/math].) Proof of the FLT in the case 2 (for example, C is divided by prime n>2, but AB not is divided) As is known, in this case the number 1°) [math]2(A+B-C)=(A+B )-(C-B )-(C-A)=[(c^n):n]-a^n-b^n=2abc[/math]. From the other side, according to lemma 1, the number 2°) [math](A+B )n-(C-B )-(C-A) [/math], or [math]c^n-a^n-b^n=2c[/math], since the dividers of the number ab are not the dividers of the number [math]c^n-a^n-b^n[/math]. Сравнивая числа [math][(c^n):n]-a^n-b^n=2abc[/math] и [math]c^n-a^n-b^n=2c[/math], что увеличение левой части равенства приводит к уменьшению правой части равенства. И мы имеем противоречие. Таким образом, оба случая ВТФ доказаны. Comparing the equalities [math][(c^n):n]-a^n-b^n=2abc[/math] and [math]c^n-a^n-b^n=2c[/math], we see that an increase in the left side of the equality leads to the decrease of the right side of the equality. And we have a contradiction. Thus, both cases of the FLT are proven.

Resumes In the recent days for me it was possible to find and to prove extremely important Lemma 1: The number [math]U=A+B-C[/math] does not contain the prime dividers, different from the dividers of the number abc. And now the case of the FLT, when the number ABC is not divided by n, proves exceptionally simply: From the equality [math]A^n+B^n=C^n[/math] (with relatively prime A, B, C and prime n>2) follows the contradictory equality [math]A+B-C=abc[/math] (where [math]a^n=C-B[/math], [math]b^n=C-A[/math], [math]c^n=A+B[/math]), since, according to Little Fermat's theorem, left side are divided by n, but right is not divided. I assume that the case of the FLT, when the number ABC is divided by n, proves with the aid of the hypothesis: Hypothesis: If one of the relatively prime numbers A, B, C (in the Fermat’s equality) is divided by the prime n (>2), then the number U=A+B-C contains the prime divider, which is not been the divider of the number abc, which contradicts lemma 1. Continuation follows.

Oct 16, 2010 Universal centuries-old error (Proof of the case of 2b°: [math]U=abcn^k[/math]) It is improbable, but with the relatively prime (and WHOLES!) A, B, C (and a, b, c) [with ABC (and abc) not divided by n] the number [math]A+B-C[/math] NOT CONVERTED into the form [math]abcn^k[/math], since the number [math]n^k[/math] is not the coefficient of the number abc, but its DIVIDER and the k-digit ENDING of the number [math]A+B-C[/math] (=U)! (And the absence of the integral solution of the Fermat’s equality is present. By the way, there is a second proof: in the numeration system on the prime base n>2 all three numbers A, B, C have at the end a digit 1.) Example (in the decimal system): The sum of the numbers a=123 and b=477 is equal 600 and, therefore, it is divided by 100, but it does not follow from this fact that a+b=(123+477)*100=60000!!! Thus, we reduced Fermat's last theorem to the only case: [math]A+B-C=abc[/math], where one of the numbers A, B, C (and a, b, c) has at the end k zeros. There is reason to believe that in the field of real numbers with [math] (A, B, C)>1[/math] the number [math]a+B-C>abc[/math], but we will give another proof (block diagram): It is easy to show that all significant parts of the numbers A, B, C have at the end a digit 1. And then it is possible to show: if the number U has at the end k zeros, then it has at the end and k+1 zeros. The interesting proof of this fact will be given later.

On the second case ([math]A+B-C=U=abcn^k[/math]) If one of the numbers (for example, C) finishes on k of zeros (in the base n), then the sum of other numbers ([math]A+B[/math]) finishes on kn-1 zero. Therefore the coefficient [math]n^k[/math] can be included in the number C: C=cr, where the number c finishes on k of zeros. And now the number U can be recorded in the form: 2a°) [math]U=abc[/math]. But if the number ABC is not divided by n, then 2b°) [math]U=abcn^k[/math], where the numbers a, b, c are not divided by n and, as were many times shown earlier, k>1. It is possible that case 2 can be proven so simply, as case 1. Interesting method of proof consists of the passage in the field of real numbers. Thus, for n=3 computer gives value for U equally 3abc, while theoretically U=abc (if abc it is divided by 3) and U=9abc (if abc it is not divided by 3).

Unexpected turning in the study New idea Let us assume that for relatively prime A, B, C and prime n>2 there is a equality 1°) [math]A^n+B^n=C^n[/math], where it is widely-known that [math]A+B-C=U>0[/math]. Then two cases are possible: Case 1: U=us, where the number ABCn is not divided by prime s. Case 2: s=1 and, therefore, [math]U=abcn^k[/math], where (according to Fermat’s little theorem) k>0 and numbers a, b, c are the greatest dividers respectively in the pairs of numbers (A and C-B ), (B and C-A), (C and A+B ). (Two of the numbers of C-B, C-A and A+B - for example, C-B and C-A - they are n-th degrees.) Case 1. The impossibility of case 1 directly follows [in the numeration system on the base s with the sum of remainders a’, b’, c’ (with the surplus or with the deficiency) or [math]a’+b’-c’=s[/math]] from the simple lemma: If [math]a+b-c=s[/math] and [math]0<(a, b, c)<s[/math], then the number [math]a^n+b^n-c^n[/math] is not divided by s. Here n and s are prime. 2. Case [math]A+B-C=U=abcn^k[/math], where k>0 (third case there does not exist) it is decomposed into two subcases: 2a. The number C is divided by n; 2b. The number C (and AB) C is not divided by n; * * * Continuation follows.

One additional attempt with the same idea and with the same tools Proof is divided off into three completely similar cases: 1) AB [2) AC and 3) BC] not divided by prime n>2. Let us take Fermat’s equality [case 1] for relatively prime A, B, C in the form 1°) [math]A^n=(C-B )P=(C-B )p^n=a^n*p^n[/math], and also [math]B^n=(C-A)Q=(C-A)q^n=b^n*q^n[/math], [math]C^n=(A+B )R[/math]. Let us take the prime divider [math]m=tn+1[/math] of the number S from the equality [math](C-B )+(C-A)=a^n+b^n=(a+b)S=2C-(A+B )[/math], which is not been the divider of the number A+B. (It is not difficult to show that this divider exists; if it is required, this will be shown later.) Then, as follows from the fermat's theorem, the number 2°) [math]A^{tn}-B^{tn}[/math] [just as the number [math](C-B )^{tn}-(C-A)^{tn}[/math]] is divided by m. However, quick calculation shows that the number [math]A^{tn}-B^{tn}[/math] is not divided by m. It would be a good thing of this to be convinced.

Good-bye, Fermat! Classical proof of Fermat's last theorem Actually school proof of the FLT is finally found! (Simplest logical substantiations are omitted.) Let us take Fermat's equality in the form: 1°) [math]A^n=(C-B)P=a^n*p^n[/math] [and [math]B^n=(C-A)Q=b^n*q^n[/math], [math]C^n=(A+B)R[/math]], where 2°) prime n is more than 2, the numbers A, B, C relatively prime, AB not is divided by n [the 1st case], 3°) r - prime divider of the number R of the form [math]r=tn+1[/math] (which, as is known, exists and, moreover, the numbers A+B and A-B is not divided by r). Then, according to Fermat's little theorem, they are divided by r the numbers: [math](C-B )^{tn}-1[/math], therefore, and 4°) [math]B^{tn}-1[/math] and [math]p^{tn}-1[/math], or [math]P^t-1[/math], therefore, and 5°) [math]B^{t(n-1)}-1[/math], therefore, and [math][b^{tn}-1]-[b^{t(n-1)}-1][/math], or [math]B^{t(n-1)}*(B-1)[/math], therefore, and 6°) B-1, и, согласно рассуждениям, аналогичным в п.п. 4-5°, 7°) A-1, следовательно, и 8°) (A-1)-(B-1), или A-B. 6°) B-1, and, according to the reasonings, analogous into 4-5°, 7°) A-1, therefore, and 8°) (A-1)-(B-1), or A-B. But, as it is known (and which it was many times shown by the author on the mathematical forums), the number A-B is not divided by prime r. Cases, when the number AC (or BC) is not divided by n, prove analogously. Thus, Fermat's last theorem is proven for all odd degrees. On October 7 to 8, 2010 Good-bye, Fermat! Classical proof of Fermat's last theorem Actually school proof of the FLT is finally found! (Simplest logical substantiations are omitted.) Let us take Fermat's equality in the form: 1°) [math]A^n=(C-B)P=a^n*p^n[/math] [and [math]B^n=(C-A)Q=b^n*q^n[/math], [math]C^n=(A+B)R[/math]], where 2°) prime n is more than 2, the numbers A, B, C relatively prime, AB not is divided by n [the 1st case], 3°) r - prime divider of the number R of the form [math]r=tn+1[/math] (which, as is known, exists and, moreover, the numbers A+B and A-B is not divided by r). Then, according to Fermat's little theorem, they are divided by r the numbers: [math](C-B )^{tn}-1[/math], therefore, and 4°) [math]B^{tn}-1[/math] and [math]p^{tn}-1[/math], or [math]P^t-1[/math], therefore, and 5°) [math]B^{t(n-1)}-1[/math], therefore, and [math][b^{tn}-1]-[b^{t(n-1)}-1][/math], or [math]B^{t(n-1)}*(B-1)[/math], therefore, and 6°) B-1, и, согласно рассуждениям, аналогичным в п.п. 4-5°, 7°) A-1, следовательно, и 8°) (A-1)-(B-1), или A-B. 6°) B-1, and, according to the reasonings, analogous into 4-5°, 7°) A-1, therefore, and 8°) (A-1)-(B-1), or A-B. But, as it is known (and which it was many times shown by the author on the mathematical forums), the number A-B is not divided by prime r. Cases, when the number AC (or BC) is not divided by n, prove analogously. Thus, Fermat's last theorem is proven for all odd degrees. On October 7 to 8, 2010 ++++++++++++++++++ Unfortunately (as always - small), in the fifth point is contained school error, and therefore as a whole proof is erroneous. It was impossible to remove error. Only useful result: all prime dividers of the form tn+1 of the numbers A, B, C in the base n finish by 01. Erroneous proof leaves for the archive.

Passed proof Earlier this version of proof it was rejected on the misunderstanding. Idea of the proof: if (with relatively prime integers A, B and C) [math]A^n+B^n=C^n[/math], then the numbers A-B and R in the equality [math]A^n-B^n=(A-B)R[/math] are not relatively prime, that is impossible. Tthe proof of the FLT is based on the simple lemma, repeatedly proven earlier on the forums. Lemma 1. with relatively prime integers a and b and a-b not multiple prime n>2 each prime divider r of the number R in the identity [math]a^n-b^n=(a-b)R[/math] takes the form: [math]r=pn+1[/math]. Proof of Fermat's last theorem Let us assume that with relatively prime integers A, B and C and (let us assume) AB not multiple prime n>2 there is a equality 1°) [math]A^n+B^n=C^n[/math], where, as is known, 2°) [math]C-B=a^n[/math]; [math]C-A=b^n[/math]; [math]A^n-B^n=(A-B)R[/math]; [math]A-B=a^n-b^n[/math]; the numbers [math]C, A-B[/math] and [math]R[/math] are relatively prime. Let us take the prime divider r of the number R of the form [math]r=pn+1[/math]. Then two numbers, namely: 3°) [math]A^n-B^n[/math] [[math]=(A-B)R[/math]] and, according to fermat's little theorem, [math](C-B )^{pn}-(C-A)^{pn}[/math] [[math]=(A-B)T[/math]], and means the numbers R and T, they are divided into r. However, as we will attempt to show below, the numbers R and T are relatively prime. (Continuation it follows).