Victor Sorok

Here are the minimum requirements for the lemma, necessary for the brief proof of the FLT: "Lemma. If integers a, b, c have only one common divisor 1, a+b=c and numbers (c^n-a^n) and (c^n-b^n) have common divisor d>2, then with n>1 one of the numbers c-b, c-a, a+b, c+b, c+a, a-b is divided by d ".

Interesting lemma If P.Ferma knew the proof of the following lemma "Lemma. If numbers [MATH]a, b, c[/MATH], have no common factor, numbers [MATH]c-a[/MATH] and [MATH]c-b[/MATH] are also mutually-prime and n is odd, then the numbers [MATH](c^n-a^n)/(c-a)[/MATH] and [MATH](c^n-b^n)/(c-b)[/MATH] are also mutually-prime", then with its aid it is possible to briefly and simply prove Fermat's last theorem. Actually, in the Fermat’s equality (where numbers [MATH]a, b, c[/MATH] have no common factor and n is odd) numbers [MATH]c-a[/MATH] and [MATH]c-b[/MATH], obviously, mutually-prime. And then from the Fermat's little theorem it follows that with prime [MATH]q>2c[/MATH] the numbers [MATH]c^{q-1}-a^{q-1}[/MATH] and [MATH]c^{q-1}-b^{q-1}[/MATH] are multiple by q. And since, according to lemma, the numbers [MATH] (c^{q-1}-a^{q-1})/(c-a)[/MATH] and [MATH] (c^{q-1}-b^{q-1})/(c-b) are mutually-prime (i.e. have no common factor), then one of the numbers [MATH]c-a[/MATH] and [MATH]c-b[/MATH] is divided by [MATH]q (>2c>c-b>c-a)[/MATH], i.e., the solution of the Fermat’s equation is not integer. It remains to learn, who and when proved lemma.

Statement «If a and b is not divide by 3, a+b is divide by 3 and odd n=3t+2, then (an+bn)/(a+b)=3p+1» is wrong. I return to binary system and take long time-out. V.S.

Answers to the questions 1) I think there is a mistake here: please take[MATH]a=5, b=7, c=11[/MATH] then [MATH]a+b, c+b, c-a [/MATH] are divided by [MATH]3[/MATH]. what do you say? Moshe A good counterexample, but it does not refer to Fermat’s equality: In equality [MATH]aa^{n-1}+bb^{n-1}-cc^{n-1}=0[/MATH] only one of the numbers [MATH]a, b, c[/MATH] can be divided by [MATH]3[/MATH], and if [MATH]a[/MATH] is not divided by [MATH]3[/MATH], then [MATH]c-b[/MATH] is not divided by [MATH]3[/MATH]. Furthermore, number [MATH]a+b-c[/MATH] is divided by [MATH]3[/MATH], since all last digits in numbers [MATH]a^{n-1}, b^{n-1}, c^{n-1}[/MATH] (if abc is not divided by [MATH]3[/MATH]) are [MATH]1[/MATH]. Victor

BUT... The proof of the FLT for odd n=3t+1 will be represented later. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Elementary proof of Fermat's last theorem in base [MATH]3[/MATH] for the odd [MATH]n=3t+2[/MATH] ([MATH]t=1, 2, …,[/MATH] numbers [MATH]a, b, c[/MATH] mutually-prime). It is easy to see that also numbers [MATH]a+b, c-b, c-a[/MATH] have no common divisors. It is evident from the simple calculations that among six mutually- prime numbers [MATH]a+b, c-b, c-a, a-b, c+b, c+a[/MATH] one and only one is divided by [MATH]3[/MATH]. Proof uses the lemma: 1°. If numbers c and a do not have common divisors, prime [MATH]d[/MATH] is a divisor of number [MATH]c-a[/MATH] and [MATH]d[/MATH] is not a divider of number[MATH]q>1[/MATH], then number [MATH]P=(c^q-a^q)/(c-a) [/MATH] is not divided by [MATH]d[/MATH]. *** Case 1: [MATH]c-a=d3^k[/MATH], where [MATH]d[/MATH] is not divided by [MATH]3[/MATH] and [MATH]k>0[/MATH]. Then the numbers [MATH](c^n-a^n)/(c-a) [/MATH] and [MATH]c^{n-1}[/MATH] (or [MATH]a^{n-1}[/MATH]) have on the ends the digit [MATH]1[/MATH], and number [MATH]D=(c^n-b^n)/(c-b)-c^{n-1}=a(c^{n-2}+… +a^{n-2})=aP[/MATH] must finish on [MATH]0[/MATH] (i.e. to be divided by [MATH]3[/MATH]). But number [MATH]D[/MATH] does not finish by [MATH]0[/MATH], since neither [MATH]a[/MATH] nor [MATH]P[/MATH] in number [MATH]c^{n-1}-a^{n-1}[/MATH] (see lemma) they are not divided by [MATH]3[/MATH]. Case 2: [MATH]c+a=d[/MATH], where d is not divided by [MATH]3[/MATH]. Proof is analogous, but here [MATH]D=(c^n+b^n)/(c+b)[/MATH]. =============== The proof of the FLT for odd [MATH]n=3t+1[/MATH] will be represented later.

I had lost. But cf. Splendid joke. Thanks

Excuse me last time, please! Splendid joke. Well known that: If in the equality 1°. a^n+b^n=c^n the numbers a, b, c have no common divisors, then the number 2°. a+b-c=un^2. It is obvious also, that if abc does not divided by n, then the number 3°. a^3n+b^3n-c^3n=Un^2. From here it follows a simple proof of the Fermat’s Last Theorem (at least, for case when abc does not divided by n): Let us use the identity for A+B=C: 4°. A^3+B^3-C^3=3ABC. But the number a^3n+b^3n-c^3n=3(a^3n)(b^3n)(c^3n) (cf. 4°) does not divided entirely by n^2 (cf. 3°)!!!

It's error. But: Last day for publication of elementary proof of the FLT. Here is the most perspective from my tentatives in base n. The numbers a, b, c have no common factors, n is prime, n>2. There is very simple proof of FLT with help following lemma. Lemma (hypothesis). The number D = a^n+b^n-c^n, where a+b-c=u=u’n^k (i.e. u has k zeros on the end, k>1), is presented in the form: D = nabcd, where d has no divisor n. (Apropos, if a+b-c=0, то a^3+b^3-c^3=3abc.) Proof of FLT. 1°. Let (E =) a^n+b^n-c^n = 0, where 2°. a+b-c=u=u’n^k, i.e. u has k zeros on the end. And 3°. (c-b)+(c-a)-(c+b)=-2u’n^k. Then according to 1° 4°. D-E= [(c-b)^n+(c-a)^n-(c+b)c^n] - (a^n+b^n-c^n)= =[(c-b)^n - a^n]+[(c-a)^n - b^n]-[(c+b)^n - c^n]=(c-a-b)(p+q+r)=-u(p+q+r)=u”n^(k+1), where the numbers p, q, r have one zero on the ends, and u has k zeros on the end. But according to Lemma 5°. D= (c-b)^n+(c-a)^n-(c+b)c^n = n(c-b)(c-a)(c+b)d. And we have a contradiction.

Unfortunately, now I do not remember the proof of Lemma: "If a+b-c=dn^k (d do not divide by n) and the number a, b, c have no commmon divisors, then a^n+b^n-c^n=d’n^(k+1)". BUT… 1) For two numbers (a and b) a proof very simple (I can show it). 2) If a+b-c=dn^k, то (a+b)-(c-b)-(c-a)=2dn^k and easy to show (I can show it), that (a^n+b^n-c^n) + [(a+b)^n-(c-b)^n-(c-a)^n] = fn^(k+1). 3) For 15 yars anybody did show, that Lemma is not true. 4) If the proof of FLT rest on Lemma only, then I try to find proof in the next time. Victor

Truly, number of divisors in numbers D and E is different! However it is necessary to introduce amendments into previous proof: It is necessary to take prime divisor n in place 2. Here revised text of the proof. Let E = a^n+b^n-c^n = 0 (Fermat’s equality), where the number a+b-c=dn^k (i.e. it has k zeros on the end). Let us introduce the numbers: A= dn^(k+1)-a, B= dn^(k+1)-b, = dn^(k+1)-c. It is obvious, that A+B-C=d’n^k (i.e. A+B-C has k zeros on the end), and (in accord to well-known lemma) the number D = D-0 = A^n+B^n-C^n has k+1 zeros on the end. BUT at the same time the number A^n+B^n-C^n = D-0 = D-E has k+2 zeros on the end. Truly, D-E = A^n+B^n-C^n - a^n+b^n-c^n = (A^n-a^n)+(B^n-b^n)-(C^n-c^n) = (A-a)P+(B-b)Q-(C-c)R = Pdn^(k+1)+Qdn^(k+1)-Rdn^(k+1) = (P+Q-R)dn^(k+1), where each from numbers P, Q, R has one 1 zero on the end. And we have a contradiction.

It is seemed that I found the simple proof of FLT in the binary system. Here it they be: Numbers D = (a+b)^n-(c-b)^n-(c-a)^n and E = a^n+b^n-c^n + D (or -D) have DIFFERENT parities. The parity of number D is a dual parity even number (from a, b, c). But the parity of number E has another value (with specific grouping of the members of sum). Detailed calculations will be published in proportion to their readiness.

Fermat's Last Theorem by Victor Sorokine. The version for n=3. LaTex. 2 mars 2006 TOOLS: Designation: [MATH]a_{(k)} [/MATH] is [MATH]k[/MATH]'digits-ending (a number ) in the number [MATH]a[/MATH] in prime base [MATH]n > 2[/MATH]. Example: [MATH]a = 3401[/MATH]: [MATH]a_{(3)} = 401[/MATH]. [MATH]a_k[/MATH] is [MATH]k[/MATH]-th digit in the number [MATH]a[/MATH]' date=' [MATH']a_1[/MATH] ≠ [MATH]0[/MATH]. Example: [MATH]a = 3401[/MATH]: [MATH]a_3 = 4[/MATH]. Well-known lemmas: Lemma 1*. If [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH] and [MATH](c-b)_{(k)}=0[/MATH], then [MATH](c^n -b^n)_{(k+1)}=0[/MATH], and if [MATH](c^n-b^n)_ {(k+1)} = 0[/MATH] and [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH], then [MATH](c-b)_{(k)}=0[/MATH] and [MATH]R_1=0[/MATH], [MATH]R_2[/MATH] ≠ [MATH]0[/MATH], where [MATH]R=\frac{c^n-b^n}{c-b}[/MATH]. [Thus, if [MATH]r[/MATH] [[MATH]= c-b[/MATH]] is divided by [MATH]n[/MATH], then the number [MATH]R[/MATH] contains only one factor [MATH]n[/MATH] (if, of course, the digit [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH]), or: [MATH]R_1 = 0[/MATH] and [MATH]R_2[/MATH] ≠ [MATH]0[/MATH]. This fact is easy proved with grouping of members of the number [MATH]R[/MATH] in pairs with following separation of the factor [MATH](c - b)^2[/MATH] in each pair.]2* Лемма. Если [MATH]a_1[/MATH] ≠ [MATH]0[/MATH] и [MATH]k > 0[/MATH], тогда существует такое [MATH]d[/MATH], что [MATH](ad)_{(k)} = 1[/MATH]. Lemma 2*. If [MATH]a_1[/MATH] ≠ [MATH]0[/MATH], [MATH]a_{(2)}=(a^n)_{(2)}[/MATH], and [MATH]k > 0[/MATH], then there exsists such [MATH]d^{nn}[/MATH] that [MATH](ad^nn)_{(k)}=1[/MATH]. Lemma 3*. If the numbers [MATH]c[/MATH] and [MATH]b[/MATH] have no common factors and the number [MATH]r [= c-b] [/MATH] is not divided by [MATH]n[/MATH], then the numbers [MATH]r[/MATH] and [MATH]R[/MATH] have no common factors. Proof of FLT for particular case: n=3 and k=2 (1°) Let [MATH]a^3=c^3-b^3=rR[/MATH], [MATH]a_1[/MATH] ≠ [MATH]0[/MATH], [MATH]a[/MATH], [MATH]b[/MATH], [MATH]c[/MATH] have no common factors, therefore the numbers [MATH]r=(c-b)[/MATH] and [MATH]R = (c^3-b^3)/(c-b)=c^2+cb+b^2=(c-b)^2+3cb[/MATH] have no common factors and (2a°) [MATH]r = (c-b)= r'^3[/MATH], (2b°) [MATH]R = c^2+cb+b^2 = R'^3=(c-b)^2+3cb=(r'^3)^2+3cb=(r'^2)^3+3cb[/MATH], (2с°) [MATH]u = a + b-c[/MATH], where [MATH]u_{(2)} = 0[/MATH], the digit [MATH]u_{3}[/MATH] ≠ [MATH]0[/MATH], [MATH]k > 0[/MATH] (corollary from 1° and from Fermat's Little theorem). (3°) [MATH][(c-b)^{(n-1)} - R]_{(k+1)}=0[/MATH] [KEY of the proof!], since [MATH](k+1)[/MATH]-digits endings in the numbers: [MATH](c-b)^n-(c-b)R[/MATH], [MATH](c-b)^n-a^n[/MATH], [MATH][(c-b)-a]Q[/MATH], [MATH]uQ[/MATH] are equal [MATH]0[/MATH], since [MATH]u_{(k)} = 0[/MATH] (cf. 2c°) и [MATH]Q_1 = 0[/MATH] (cf. 1*). Case 1: [MATH](abc)_1[/MATH] ≠ [MATH]0[/MATH]. Since in 2b° the endings [MATH]R'_{(2)}= r'^2[/MATH][MATH]_{(2)}=1[/MATH], then the endings [MATH]R'^n[/MATH][MATH]_{(2)}= (r'^2)^3[/MATH][MATH]_{(2)}=01[/MATH], and since the digit [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH], then the equality [MATH]R'^3=(r'^2)^3+3cb[/MATH] is IMPOSSIBLE. Случай 2: [MATH]b_1=0[/MATH], but [MATH](ac)_1[/MATH] ≠ [MATH]0[/MATH]. According to key equality 3°, the ending [(c-b)^2-R}_{(3)}=0. This fact does not change from the conversion [MATH]3[/MATH]-digits ending of the number [MATH]a[/MATH] into [MATH]1[/MATH] (with the aid of the multiplication of Fermat's equality by a certain number [MATH]d^9[/MATH] with the retention of properties 2а° - 2c° - cf. 2*; for the clarity of the perception, the designations of letters remain previous, but TITLE, and asterisk is assigned to numbers r and R. Now the numbers [MATH]C[/MATH] and [MATH]A[/MATH] finish to 001. And if [MATH]k=2[/MATH], then the number [MATH]U[/MATH] (and the number [MATH]B[/MATH] - since the number [MATH]C-A[/MATH] it finishes even to 5 zeros - cmf. 1*) as before finishes to 2 zeros. But from the direct calculation of the ending of the number [MATH]R^*-(C-B)^2[/MATH] we see that the third number (from the end) in this number IS NOT EQUAL to zero: second term in [MATH]R^*[/MATH] is [MATH]B[/MATH], and one in [MATH](C-B)^2[/MATH] is [MATH]2B[/MATH], therefore third digits in [MATH]R^*-(C-B)^2[/MATH] is equal to [MATH](B-2B)_3=(-B)_3[/MATH], where [MATH]B_3[/MATH] ≠ [MATH]0[/MATH]. Thus, the second case consists of the following assertions: 1) key conclusion 3°: the number [MATH]r^2-R[/MATH] finishes to [MATH]3[/MATH] zero. 2) [MATH]3[/MATH]-digits ending of the number [MATH]a[/MATH] can be converted into 1 with the aid of a certain coefficient of the equality 1° of kind [MATH]d^9[/MATH]. 3) [MATH]3[/MATH]-th digits in the numbers [MATH]r^{*2}[/MATH] and [MATH]R^*[/MATH] are different from zero and they are not equal to each other, and therefore 4) in their difference [MATH]3[/MATH]-th digit IS NOT EQUAL to zero, i.e. "the number [MATH]r^{*2}-R^*[/MATH] finishes ONLY to [MATH]2[/MATH] zeros"! But now you will compare it with point 1). Now the question: in which of the 4 points is contained the error? Or at least: which of the 4 points does cause doubt? __________________ The proof of the second case is generalized to any prime [MATH]n>2[/MATH] and [MATH]k>0[/MATH] without any additional operations. But precisely this case is considered in the VTF-theory as the most difficult for the proof, since it disrupts "symmetry", "order": one of three "r" is not a power. However, not this case enraptured P.Ferma. The proof of the first case in general form awaits its publication. Its salt - very spectacular (it would seem, impossible!) the algebraic conversion of the formula of the number R, after which the zero ending of the R becomes obvious.

Fermat's Last Theorem by Victor Sorokine. The simplest version TOOLS: Designation: [MATH]a_{(k)} [/MATH] is [MATH]k[/MATH]'digits-ending (a number ) in the number [MATH]a[/MATH] in prime base [MATH]n > 2[/MATH]. Example: [MATH]a = 3401[/MATH]: [MATH]a_{(3)} = 401[/MATH]. [MATH]a_k[/MATH] is [MATH]k[/MATH]-th digit in the number [MATH]a[/MATH], [MATH]a_1[/MATH] ≠ [MATH]0[/MATH]. Example: [MATH]a = 3401[/MATH]: [MATH]a_3 = 4[/MATH]. Well-known lemma: Lemma 1*. If [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH] and [MATH](c-b)_{(k)}=0[/MATH], then [MATH](c^n -b^n)_{(k+1)}=0[/MATH], and if [MATH](c^n-b^n)_ {(k+1)} = 0[/MATH] and [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH], then [MATH](c-b)_{(k)}=0[/MATH] and [MATH]R_1=0[/MATH], [MATH]R_2[/MATH] ≠ [MATH]0[/MATH], where [MATH]R=\frac{c^n-b^n}{c-b}[/MATH]. [Thus, if [MATH]r[/MATH] [[MATH]= c-b[/MATH]] is divided by [MATH]n[/MATH], then the number [MATH]R[/MATH] contains only one factor [MATH]n[/MATH] (if, of course, the digit [MATH](cb)_1[/MATH] # [MATH]0[/MATH]), or: [MATH]R_1 = 0[/MATH] and [MATH]R_2[/MATH] # [MATH]0[/MATH]. This fact is easy proved with grouping of members of the number [MATH]R[/MATH] in pairs with following separation of the factor [MATH](c - b)^2[/MATH] in each pair.] The proof of FLT (1°) Let [MATH]a^n=c^n-b^n=rR[/MATH], where [MATH]n[/MATH] is prime, [MATH]a_1[/MATH] ≠ [MATH]0[/MATH], [MATH]a[/MATH], [MATH]b[/MATH], [MATH]c[/MATH] have no common factors, then (2a°) [MATH]r=c-b= r'^n [/MATH] and [MATH]R=\frac{c^n-b^n}{c-b}=R'^n [/MATH], [MATH]a = r'R'[/MATH]; (2b°) [MATH]u=a+b-c[/MATH], where [MATH]u_{(k)}=0[/MATH], the digit u_{k+1}[/MATH] ≠ [MATH]0[/MATH], [MATH]k > 0[/MATH] (corollary from 1° and Little Theorem). *** (3°) The [MATH](k+1)[/MATH]'digits-endings in the equivalent numbers [MATH](c-b)^n-a^n[/MATH], [MATH](c-b)^n-(c-b)R[/MATH], [MATH](c-b)^n-a^n[/MATH], [MATH][(c-b)-a]Q[/MATH] (cf. 1*), [MATH] uQ[/MATH] are equal to [MATH]0[/MATH], since [MATH]u_{(k)}=0[/MATH] (cf. 2b°) and [MATH]Q_1=0[/MATH] (cf. 1*). (4°) From here we have: [MATH]R_{(k+1)}=[(c-b)^{n-1}][/MATH][MATH]_{(k+1)} = (r'^n)^{n-1}[/MATH][MATH]_{(k+1)} = (r'^{n-1})^n[/MATH][MATH]_{(k+1)}[/MATH]. [KEY of the proof!] Let's compare the equality [MATH]a = r'R'[/MATH] (cf. 2a°) with [MATH](k+1)[/MATH] 'digits-endings: (5°) [MATH]a_{(k+1)} = (r'R')[/MATH][MATH]_{(k+1)} =[/MATH]... (cf. 4°) …[MATH]=(r'r'^{n-1})[/MATH][MATH]_{(k+1)} = (r'^n)[/MATH][MATH]_{(k+1)} =[/MATH]… (cf. 2a°) …[MATH]= (c-b)[/MATH][MATH]_{(k+1)} [/MATH], that contradicts to (2b°). The FLT is proved. Victor Sorokine

Situation as of on 10 February 2006. Thus, judging by the set it is indicative, the elementary proof of great theorem is found. After the demonstration of the simplest key formula of (4°) the proof appears simpler than the Pythagorean theorem. It is easy to explain, why in 300 years no one came close to formula (4°): with the derivation of formula it's used the expression [MATH](c-b)^n[/MATH], which is absent in the Fermat's equality. The very detailed discussion passed on the forum lib.mexmat.ru (with Mr. Someone), which made it possible to in detail explain each assertion. However, complete understanding with the opponent isn't reached. The analysis of difference can prove to be, in my view, very instructive for understanding of the finesses of mathematics. Mr. Someone attempted to refute my proof with the aid of the numerical counterexample, in which complete Fermat's equality is substituted with partial equality at the [MATH]10[/MATH]'digits-endings of powers. I.e., Mr. Someone attempts to refute equality with the aid of the replacement of equality into the inequality. And here to understand this logical error my opponent can't. The fact is that the replacement of equality into the inequality conducts to THE ESSENTIAL loss of information. So if two odd powers are equal ON ALL DIGITS, then from equality [MATH](k+1)[/MATH]-th digits in the powers follows equality [MATH](k+1)[/MATH]-th digits in the bases. But the consequence indicated it weakens during the replacement of complete equality by equality at the [MATH](k+1)[/MATH]'digits-endings: now (see lemma 1°) if two powers are equal at the [MATH](k+1)[/MATH]'digits-endings, then their bases are deliberately equal on k-th digits, and here equality at the [MATH](k+1)[/MATH]'digits-endings can not be carried out. But since in the counterexample (not entirely adequate, since the right side of the equality is not a power) the equality of the left and right sides of the Fermat's equality is carried out on [MATH](k+2)[/MATH]'digits-endings, then allegedly it follows from this that the bases are equal only at the [MATH](k+1)[/MATH]'digits-endings. The replacement of complete Fermat's equality to the equality only at the endings leads to the impossibility to show the key relationships of 4°-5° and as this consequence to see inequality [MATH](k+1)[/MATH]-th digits in two parts of equality (1°). In connection with this we ended our dialogue. However, all the formulas, given in the proof, weren't disproved. Was refuted only their interpretation. Most of all Mr. Someone attempted to show that the [MATH](k+1)[/MATH]'digits-ending of the right side of the Fermat's equality (1°) is not represented in the form of multiplication some n of the numbers with the equal [MATH](k+1)[/MATH]'digits-endings. In particular, it did not agree to recognize the [MATH](k+1)[/MATH]'digits-ending of the number [MATH]R[/MATH] (equal to 1) equal to multiplication [MATH]n-1[/MATH] of ones - in spite of the fact that the number of simple cofactors in the number [MATH]R[/MATH] deliberately more [MATH]n-1[/MATH]. This acknowledgement would indicate the acknowledgement of proof as accurate. And therefore I again will dwell on this very "difficult" moment of the proof. Thus, according to conversion the [MATH](k+1)[/MATH]'digits-ending of number [MATH]c-b[/MATH] into [MATH]1[/MATH], identity 5° and strict concept power, all simple cofactors in the right (as in the leftist) part of the equality can be grouped in n of "great" cofactors with THE EQUAL [MATH](k+1)[/MATH]'digits-endings and with [MATH](k+1)[/MATH]-th digit equal to [MATH]1[/MATH] (more precise: [MATH]000…0001[/MATH]). I by no transfer of simple cofactors of one "great" cofactor in another - WHEN the [MATH](k+1)[/MATH]'digits-endings of all "great" cofactors will be equal to each other - to obtain other [MATH](k+1)[/MATH]'digits-endings in "great" cofactors is impossible, since their multiplication CANNOT CHANGE from the transposition of cofactors. This is the fundamental axiom of arithmetic, and it is foolish to attempt it to refute - moreover only for this reason not recognizing as accurate the simple proof FLT. (I note that the proof contains not united calculation - all formulas are well-known.) I will give the detailed proof of "difficult" place itself. Final conclusions from identity 5°: Let [MATH]p_i[/MATH] - prime cofactor of the left side of equality 1° (number [MATH]a^n[/MATH]), [MATH]q_i[/MATH] - simple cofactor of the left side of equality 1° (number [MATH]rR[/MATH]), [MATH]p'_i[/MATH] - [MATH](k+1)[/MATH]'digits-ending of number [MATH]p_i[/MATH], [MATH]q_i[/MATH] - [MATH](k+1)[/MATH]'digits-ending of number [MATH]q_i[/MATH]. Then 1) set {[MATH]p_i[/MATH]} = {[MATH]q_i[/MATH]} (consequence of equality 1°); consequently: 2) set {[MATH]p'_i[/MATH]} = {[MATH]q'_i[/MATH]} (consequence of the uniqueness of the idea of the number in prime base n); consequently: 3) set {[MATH]p'_i[/MATH]}/n = {[MATH]q'_i[/MATH]}/n (as equal parts of the equal); consequently: 4) work [MATH]P'[/MATH] all numbers of {[MATH]p'_i[/MATH]}/n is equal to work [MATH]Q'[/MATH] all numbers of {[MATH]q'_i[/MATH]}/n; consequently: 5) [MATH]P'_{(k+1)} = Q'_{(k+1)}[/MATH] (consequence of equality[MATH]P'= Q' [/MATH]). Final conclusion. We have a contradiction: k+1-th digit on the left side of equality (1°) [MATH]p'_{k+1}[/MATH] ≠ [MATH]0[/MATH], and THE SAME digit in right side of [MATH]Q'_{k+1} = 0[/MATH].

1) For n=1 the number u [=a+b-c] has no non-zero digit and R=1. My proof doesn't work. [And cf. Designation: n > 2.] 2, 3) If k [or k+1] is prime my proof works without hardship. V.S.

IMPORTANT FORMALITY: in place of: (5°) From [MATH]n-1[/MATH] possible values of the ending [MATH](c-b)_{(k+1)} [/MATH] only one satisfies to the equation 1°, namely: [MATH](c-b)_ {(k+1)}=1[/MATH]. must have: (5°) If [MATH](c-b)_ {(k+1)}=1[/MATH], then : [MATH][(c-b)^{n-1}][/MATH][MATH]_{(k+1)}= [(c-b)_{(k+1)}][/MATH][MATH]^{n-1}[/MATH].

Fermat's Last Theorem: All digits in a+b-c are zeros TOOLS: Designation: [MATH]a_{(k)} [/MATH] is [MATH]k[/MATH]'digits-ending (a number ) in the number [MATH]a[/MATH] in prime base [MATH]n > 2[/MATH]. [MATH]a_k[/MATH] is [MATH]k[/MATH]-th digit in the number [MATH]a[/MATH], [MATH]a_1[/MATH] ≠ [MATH]0[/MATH]. Well-known lemmas: 1* Lemma. If [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH] and [MATH](c-b)_{(k)}=0[/MATH], then [MATH](c^n -b^n)_ {(k+1)}=0[/MATH], and if [MATH](c^n-b^n)_ {(k+1)} = 0[/MATH] and [MATH](cb)_1[/MATH] ≠ [MATH]0[/MATH], then [MATH](c-b)_{(k)}=0[/MATH] and [MATH]R_1=0[/MATH],[MATH] R_2[/MATH] ≠ [MATH]0[/MATH], where [MATH]R=\frac{c^n-b^n}{c-b}[/MATH]. 2* Lemma. If [MATH]a_1[/MATH] ≠ [MATH]0[/MATH] and [MATH]k > 0[/MATH], then there exsists such [MATH]d[/MATH] that [MATH] (ad)_{(k)}=1[/MATH]. [Optional Lemma. [MATH]a^n_{(k)} [/MATH] is having a single meaning from [MATH]a_{(k-1)}[/MATH], but [MATH]a^n_{(k)} [/MATH] is not a function from [MATH]a_{(k)} [/MATH].] The proof of FLT (1°) Let [MATH]a^n=c^n-b^n=rR[/MATH], where [MATH]n[/MATH] is prime, [MATH]a_1[/MATH] ≠ [MATH]0[/MATH], [MATH]r=c-b[/MATH] and [MATH]R=\frac{c^n-b^n}{c-b}[/MATH], (1a°) [MATH]u=a+b-c[/MATH], [MATH]u_{(k)}=0, u_{k+1}[/MATH] ≠ [MATH]0[/MATH], [MATH] k > 0[/MATH]. (2°) Let's transform the ending [MATH](c-b)_ {(k+2)} [/MATH] into [MATH]1[/MATH] with help of the multiplication of the equation 1° by [MATH]d^n[/MATH] from 2*. Then (cf. 1a°) [MATH]a_{(k)}=0[/MATH], [MATH] a_{k+1}[/MATH] ≠ [MATH]0[/MATH]. Table of symbols don,'t change. *** (3°) The [MATH](k+1)[/MATH]'digits-endings in the numbers [MATH](c-b)^n-a^n[/MATH], [MATH](c-b)^n-(c-b)R[/MATH], [MATH](c-b)^n-a^n[/MATH], [MATH][(c-b)-a]Q[/MATH], [MATH] uQ[/MATH] are equal to [MATH]0[/MATH], since [MATH]u_{(k)}=0[/MATH] (cf. 1a°) and [MATH]Q_1=0[/MATH] (cf. 1*). (4°) From here we have: [MATH]R_{(k+1)}=[(c-b)^{n-1}][/MATH][MATH]_{(k+1)}=1[/MATH]. [KEY of the proof!] (5°) From [MATH]n-1[/MATH] possible values of the ending [MATH](c-b)_{(k+1)} [/MATH] only one satisfies to the equation 1°, namely: [MATH](c-b)_ {(k+1)}=1[/MATH]. And we have a contradiction in the equation 1°: in left part [MATH]a_{k+1}[/MATH] ≠ [MATH]0[/MATH] [and [MATH]a^n[/MATH][MATH]_{k+2}[/MATH] ≠ [MATH]0[/MATH]] (cf. 1°), but in the right part [MATH](c-b)_ {(k+1)}[/MATH], or [MATH]a_{k+1}=0 [/MATH] [and [MATH]a^n[/MATH][MATH]_{k+2}=0[/MATH]] (cf. 5°). Let's notice that the sets of prime factors for both parts of the equation 1° are the same. Therefore, the sets of [MATH](k+1)[/MATH]'digits-endings are the same too. There is only one way of dividing these sets into n equal subsets. Then, the [MATH](k+1)[/MATH]'digits-endings of the numbers [MATH]a[/MATH] and [MATH]c-b[/MATH] have the same sets of factors. The two numbers are the same then. However, the digits of the rang [MATH]k+1[/MATH] are different. Therefore the equation 1° is impossible. Victor Sorokine

Fermat's Last Theorem: All digits in a + b – c are zeros (reductive version) TOOLS (usual): Designation: a_(k) is k-digits ending (a number ) in the number a in prime base n > 2. a_k [or a_{k}] is k-th digit in the number a, a_1 =/ 0. 1* Lemma. If (cb)_1 =/ 0 and (c – b)_(p) = 0, then (c^n – b^n)_(p+1) = 0, and if (c^n – b^n)_(p+1) = 0 and (cb)_1 =/ 0, then (c – b)_(p) = 0 and R_1 = 0, R_2 =/ 0, where R = (c^n – b^n)/(c – b). 2* Lemma. If a_1 =/ 0 and k > 0, then there exsists such d that (ad)_(k) = 1. [3* optional Lemma. a^n_(k) is having a single meaning from a_(k – 1), but a^n_(k) is not a function from a_(k).] The proof of FLT (in theses) (1°) Let a^n = c^n – b^n = (c – b)R, where a_1 =/0 and R = (c^n – b^n)/(c – b), (1a°) u = a + b – c, u_^n = 0, u_{k+1} =/ 0 k > 0. (2°) Let's transform the (k + 2)-digit ending (c – b)_(k+2) into 1 with help of the multiplication of the equation 1° by d^n from 2*. Then a_(k) = 0, a_{k+1} =/ 0. *** (3°) The (k+1)-digit ending in the number (c – b)^n – (c – b)R = (c – b ^n – a^n = (c – b – a)Q = uQ is equal to 0, since u_(k) = 0 (cf. 1a°) and Q_1 = 0 (cf. 1*). (4°) From here we have: R_(k+1) = [(c – b)^(n – 1)]_(k+1) = 1. [KEY of the proof!] (5°) From n – 1 possible values of the ending (c – b)_(k+1) only one satisfies to the equation 1°, namely: (c – b)_(k+1) = 1. And we have a contradiction in the equation 1°: in left part a_{k+1} =/ 0 (cf. 1°), but in the right part a_^n = (c – b)_{k+1} = 0 (cf. 5°). The proof is done. Victor Sorokine

Accurate definition of the last phrase And now after the transformation of the k+1-digit ending in the number a' (with help of the multiplication of the equation 1° by some number d^nnn – cf. Lemma 2a°) we have a contradiction in the equation 1° in the k+2-th digits: (3°) in the left part: (a^n)_{k+2} ≠ 0 since a_{k+1} ≠ 0 [cf. 1a°' date= where (c – b)_(k+1) = 1, the ending u_(k) = 0, the digit u_{k+1} ≠ 0, and a_(k) = 0]; (4°) °) in the right part: (c – b)_(k+1) = R_(k+1) = (c – b)^(n – 1)_(k+1) = 00…001 (or 1), therefore [(c – b )R]_(k+2) = {[(c – b )_(k+1)]^n}_(k+2) = 00…001 (cf. 2*), from here [(c – b )R]_{k+2} = 0. The proof is done. Victor Sorokine

Complete Proof of the Fermat's Last Theorem Tools Designation: a_(k) is k-digits ending in the number a in prime base n > 2. a_1 is last digit in the number a' date=' everywhere a_1 =/ 0. 1* Lemma. If (cb)_1 ≠ 0 and (c – b)_(k) = 0, then (c^n – b^n)_(k + 1) = 0, and if (c^n – b^n)_(k + 1) = 0 and (cb)_1 ≠ 0, then (c – b)_(k) = 0. If R = (c^n – b^n)/(c – b) where (cb)_1 ≠ 0, then R_1 = 0, R_2 ≠ 0. 2* Lemma. a^n_(k) and a_(k – 1) are determine each other unambiguously. The corollary from Binomial theorem. 2a* Lemma. If a_1 =/ 0 and k > 0, then there exists such d that (ad)_k = 1. 3* Little Fermat's Theorem: If a_1 =/ 0, then a^n_1 = a_1 and a^(n – 1)_1 = 1. (1°) Let a^n = c^n – b^n = (c – b)R, where: n is prime, a, b, c have no common factors, (1a°) a + b – c = u, where: u_(k) = 0, the digit u_{k+1} ≠ 0, k > 1 (cf. 3*); and (1b°) c – b = a'^n, R = a''^n (since the numbers c – b and R have no common factors.) [b']PROOF of the FLT [/b](12 lines) (2°) The (k+1)-digits endings in the number (c – b)^n – (c – b)R = (c – b)^n – a^n = (c – b – a)Q = uQ is equal to 0, since u_(k) = 0 (cf. 1a°) and Q_1 = 0 (cf. 1*). (2a°) From here we have: R_(k+1) = a^(n – 1)_(k+1), since the digit (c – b)_1 ≠ 0 (The corollary from 3* and 1°). KEY of the proof. And now after the transformation of the k+1-digit ending in the number a' (with help of the multiplication of the equation 1° by some number d^nnn – cf. Lemma 2a°) we have a contradiction in the equation 1° in the k+1-th digits, since k+1-digit endings in the numbers (c – b), R – therefore also in the right part of the equality – are equal to 00…001 (or 1), BUT k+1-th digit in left part of the 1° is not equal to zero, since k-th digit in the number a is NOT equal to zero [cf. 1a°, were (c – b)_(k+1) = 1]. Thus all digits in the number u are zeros. But if a + b – c = 0 the Fermat's equation is impossible! The proof is done. Victor Sorokine THE END

TOOLS for schoolbodies Designation: a_(k) is k-digits ending in the number a in prime base n > 2. a_1 is last digit in the number a, a_1 =/ 0. All numbers are done in base with prime n > 2: 0, 1, 2, 10, 11, 12, 22… For a = 1012 a_1 = a_(1) = 2, a_(2) = 12, a_(3) = 012 = 12, … 1* Lemma. If (cb)_1 =/ 0 and (c – b)_(p) = 0, then (c^n – b^n)_(p + 1) = 0, and if (c^n – b^n)_(p + 1) = 0 and (cb)_1 =/ 0, then (c – b)_(p) = 0. Proof for n = 3: c^3 – b^3 = (c – b)(c^2 + bc + b^2) = = (c – b)R, where R = c^2 + bc + b^2 = = (c^2 + b^2) + bc = (c^2 – 2b c + b^2 + 2bc) + bc = (b – c)^2 + 3bc. If b and have no common factors and the number b – c divides by 3^k, then R divides by 3^1 and c^3 – b^3 divides by 3^(k+1). 2* Lemma. a^n_(k) is reciprocally having a single meaning from a_(k – 1) (here a_1 =/ 0). Proof for n = 3 and k = 2: a_(2) = a_(1) + 3(a_2); [a_(2)]^3_(2) = {[a_(1) + 3(a_2)]^3}_(2) = {[a_(1)]^3 + 3{[a_(1)]^2}[3(a_2)] + 3{[a_(1)]}[3(a_2)]^2 + [3(a_2)]^3}_(2) = {[a_(1)]^3 + (3^2)P}_(2) = {[a_(1)]^3}_(2) + [(3^2)P]_(2) = {[a_(1)]^3}_(2), since [(3^2)P]_(2) = 0. 3* Little Fermat's Theorem: If a_1 =/ 0, then a^n_1 = a_1 and a^(n – 1)_1 = 0. Therefore (a^n + b^n – c^n)_1 = (a + b – c)_1 = (a_1 + b_1 – c_1)_1 = u_1 = 0.

Fermat's Last Theorem. If a + b – c… Complete PROOF of FLT Usual TOOLS: Designation: a_(k) is k-digits ending in the number a in prime base n > 2. a_1 is last digit in the number a, a_1 =/ 0. 1* Lemma. If (cb)_1 =/ 0 and (c – b)_(p) = 0, then (c^n – b^n)_(p + 1) = 0, and if (c^n – b^n)_(p + 1) = 0 and (cb)_1 =/ 0, then (c – b)_(p) = 0. 2* Lemma. a^n_(k) is reciprocally having a single meaning from a_(k – 1). 3* Little Fermat's Theorem: If a_1 =/ 0, then a^n_1 = a_1 and a^(n – 1)_1 = 0. (1°) Let a^n = c^n – b^n = (c – b)R, where: n is prime, a, b, c have no common factors, (1a°) a + b – c = u with u_(k) = 0, u_{k+1} =/ 0, where k > 1, and (1b°) c – b = a'^n, R = a''^n. (2°) The (k+1)-digits endings in the number (c – b)^n – (c – b)R = (c – b)^n – a^n = (c – b – a)Q = uQ is equal to 0, since u_(k) = 0 (cf. 1a°) and Q_1 = 0 (cf. 1*). (2a°) From here we have: Q_(k+1) = [(c – b)^(n – 1)]_(k+1) since [(c – b)_1 =/ 0 (corollary from 1* and 1°). (3°) {[(c – b)_(k+1)]R_(k+1)}_(k+1) = ([(c – b)_(k+1)]{[(c – b)^(n – 1)]_(k+1)})_(k+1) = {[(a')_(k+1)^n_(k+2)]{[(a')^(n – 1)^n]_(k+2)}}_(k+2) = [(a')_(k+1)^^n]_(k+3) = (c – b)^n_(k+2). (4°) From [a^n – (c – b)^n]_(k+2) = 0 we have: (a + b – c)_(k+1) that contradicts to 1a°. The proof is done. Victor Sorokine

Framework of the idea (1 phrase) In the equality A^n – (C^n – B^n) = AA^(n – 1) – (C – B)R = 0 the number u = A + B – C has k zeros at the end, but the last k + 1 digits in the numbers A^(n – 1) and R one can transform into 00…001, but then the number u = A + B – C has k + 1 zeros at the end. In any case the proof merits of careful analysis. ============ Short PROOF of FLT: If a + b – c = 0 mod(n^k) and a=/ 0 mod(n), then (c – b)^(n – 1) = [(c – b)^n]/(c – b) = a^(n – 1) mod(n^(k+1)) and therefore a + b – c = 0 mod(n^(k+1)) – cf. my Forum: http://www.ivlim.ru/fox/forum/FORUM.asp?FORUM_ID=20&CAT_ID=1&Forum_Title=%C2%E5%EB%E8%EA%E0%FF+%D2%E5%EE%F0%E5%EC%E0+%D4%E5%F0%EC%E0 Happy New Year! Victor SOROKINE