Fermat's Last Theorem. If a + b – c = 0 mod(n^k), then a + b – c = 0 mod(n^(k+1))

[Victor Sorok]

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

[Victor Sorok]

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

[Victor Sorok]

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.

[Victor Sorok]

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 only the zeros.

 

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

[Victor Sorok]

 

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

[Victor Sorok]

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

[EvoN1020v]

An advice: Use latex next time. You are putting us into confusion easily.

[Victor Sorok]

An advice: Use latex next time. You are putting us into confusion easily.

 

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

[Victor Sorok]

[b']Fermat's Last Theorem: All digits in a+b-c are zeros[/b]

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].

[ed84c]

I thought fermats last theroem was that pythagorean triples dont work with cubes, or is that a complicated way of saying it?

[EvoN1020v]

Also, are you using induction method to prove Fermat's last theorem? Where you do this:

 

1) Show it's true for n=1

2) Assume it's true for n=k, kEN

3) Prove it's true for n=k+1

 

Is this what you're doing?

[Victor Sorok]

Also' date=' are you using [b']induction[/b] method to prove Fermat's last theorem? Where you do this:

 

1) Show it's true for n=1

2) Assume it's true for n=k, kEN

3) Prove it's true for n=k+1

 

Is this what you're doing?

 

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.

[Victor Sorok]

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].

[Victor Sorok]

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

[Victor Sorok]

 

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.

[Victor Sorok]

Fermat's Last Theorem by Victor Sorokine. The version for n>2. LaTex. 27 mars 2006

Today I assembled together everything' date=' which with the need relates to the proof FLT. In effect all positions either are well known to specialists in the theory of the numbers or they prove completely simply, and therefore by large part of these proofs I omit. Moreover almost everything was discussed on the mathematical forums. So that only proof itself, which consists of one paragraph and which contains practically no calculations, is new. However, the passages of some moments are completely possible, and they in the process of consideration will be included in text.

V.S.

 

I will resemble basic facts from the arithmetic of degrees and equality the farm:

 

Designations:

 

[MATH']a_i[/MATH] - [MATH]i[/MATH]-th digit in the number [MATH]a[/MATH];

[MATH]a_{(i)}[/MATH] - [MATH]i[/MATH]-digits ending (a number ) in the number [MATH]a[/MATH] in prime base [MATH]n>2[/MATH];

everywhere in the text the number [MATH]a[/MATH] is not divided by [MATH]n[/MATH];

all proofs are conducted in the numeration system with the prime base [MATH]n>2[/MATH].

 

Lemmas:

1. If the numbers [MATH]a[/MATH] и [MATH]b[/MATH] have no common factors and [MATH]c-b[/MATH] is not divided by [MATH]n[/MATH], then the numbers [MATH]R=(c^n-b^n)/(c-b)[/MATH] and [MATH]r=c-b[/MATH] have no common factors (well-known lemma).

2. If the numbers [MATH]a[/MATH] and [MATH]c[/MATH] have no common factors and [MATH]c-b[/MATH] is divided by [MATH]n[/MATH], then the number [MATH]Q=(c^n-a^n)/(c-a)[/MATH] is divided by [MATH]n^1[/MATH], but is not divided by [MATH]n^2[/MATH] (well-known lemma). (Consequently, number of zeros at the end of the number [MATH]c^n-a^n[/MATH] to one of more than them in the number [MATH]c-a[/MATH].)

3. Corollary: If [MATH]d^n[/MATH][MATH]_{(i+1)}=e^n[/MATH][MATH]_{(i+1)}[/MATH], then [MATH]d_{(i)}=e_{(i)}[/MATH].

4. [MATH]a^n_1=a_1[/MATH] (direct corollary from 8 and from Fermat's little theorem).

5. It is obvious, if [MATH](de)_{(i)}=(fg)_{(i)}[/MATH] and [MATH]d_{(i)}=f_{(i)}[/MATH], then [MATH](e)_{(i)}= (g)_{(i)}[/MATH].

6. If [MATH]e_1[/MATH] ≠ [MATH]0[/MATH], then for any [MATH]i[/MATH] there exists such [MATH]d[/MATH], that [MATH](ad)_{(i)}=e_{(i)}[/MATH] (well-known lemma).

7. Among of digits [MATH]1,2,3,…n-1[/MATH] there exists such [MATH]g[/MATH], that [MATH]g^{n-1}[/MATH][MATH]_2[/MATH] ≠ [MATH]0[/MATH] [actually, otherwise the second digit in the sum of all numbers from 1 to [MATH]n-1[/MATH] in the [MATH]n[/MATH]-power is not equal to [MATH]0[/MATH], while of the terms, equidistant from the ends of the power series it is possible to compose vapors, in which each sum finishes to TWO zeros)].

 

8. Let, [MATH]a^n-(c^n-b^n)=0[/MATH], where

9. the numbers [MATH]a, b, c[/MATH] have no common factors and [MATH]a[/MATH] is not divided by [MATH]n[/MATH].

10. Table of symbols: [MATH]r=c-b[/MATH], [MATH]R=(c^n-b^n)/(c-b)[/MATH], [MATH]u=a+b-c[/MATH].

11. The numbers [MATH]R[/MATH], [MATH]r[/MATH] и [MATH]a[/MATH] have the form: [MATH]R=R'^n[/MATH], [MATH]r=r'^n[/MATH] и [MATH]a=r'R'[/MATH] (corollary from items 1 and 8).

12. [MATH]u_1=0[/MATH] (corollary from items 4 and 8).

13. [MATH]R_1=1[/MATH] (corollary from items 4 and 12).

14. [MATH]R_{(2)}=01[/MATH] (corollary from items 11, 13 and from Binomial theorem).

15. Analogous: If [MATH]b_1[/MATH] ≠ [MATH]0[/MATH] and [MATH]c_1[/MATH] ≠ [MATH]0[/MATH], then [MATH][(c^n-a^n)/(c-a)]_{(2)}= [(a^n+b^n)/(a+b)]_{(2)}=01[/MATH].

16. But if [MATH]c_1=0[/MATH] (or [MATH]b_1=0[/MATH]), then [MATH]a+b[/MATH] (or [MATH]c-a[/MATH]) finishes to two zeros even in the case [MATH]n=3[/MATH] (cf. item 2).

17. [MATH]u_{(2)}=0[/MATH] (corollary from items 14, 15, 16 and 8).

18. [MATH]R_{(3)}=(c-b)^{n-1}[/MATH][MATH]_{(3)}[/MATH] (key formula) (the faithfulness of assertion becomes obvious after the multiplication of this equality by [ MATH]c-b[/MATH ] and from the applications of the item 5 and Corollary in the item 2).

19. [MATH]R'_{(2)}=r'^{n-1}[/MATH][MATH]_{(2)}[/MATH] (corollary from items 18 and 3).

20. [MATH]a^{n-1}[/MATH][MATH]_{(2)}=R_{(2)}=01[/MATH] (corollary from equality [MATH]aa^{n-1}=(c-b)R[/MATH] and from items 10 and 17).

 

At finally, the proof of the FLT:

 

With the aid of the multiplication of Fermat's equality by certain [MATH]d^{nn}[/MATH] (from item 6) - with the retention of the powers: [MATH]r'^n, R'^n, a^n, [/MATH] (letter designations of the numbers they remain before!) and their two-digit endings [MATH]R'^{n-1}[/MATH][MATH]_{(2)}=a^{n-1}[/MATH][MATH]_{(2)}=01[/MATH] - we convert the two-digit ending of the number [MATH]r'[/MATH] into [MATH]0g[/MATH], where the digit [MATH]g[/MATH] is such, that

21. [MATH]g^n[/MATH][MATH]_2[/MATH] ≠ [MATH]0[/MATH] (cf. item 7).

And we arrived at the contradiction:

 

on the one hand, [MATH]a^{n-1}[/MATH][MATH]_{(2)}=01[/MATH] (cf. item 20),

on the other hand - [MATH](r'R')^{n-1}[/MATH][MATH]_{(2)}= [r'^{n-1}R'^{n-1}]_{(2)}=01[/MATH], where [MATH]R'^{n-1}[/MATH][MATH]_{(2)}=01[/MATH], and [MATH]r'^{n-1}[/MATH][MATH]_{(2)}[/MATH] ≠ [MATH]01[/MATH], and therefore [MATH]a^{n-1}[/MATH][MATH]_{(2)}[/MATH] ≠ [MATH]01[/MATH] (cf. item 20).

 

Thus, FLT is proved for oll prime [MATH]n>2[/MATH].

[Victor Sorok]

Proof FLT. Final text.

 

Designations:

a_(i) – the i-th digit from the end in the number a in the prime base n>2.

a_(p, r) – the number, comprised of the digits of the number a from the rank p to the rank r.

9 – designation of digit n-1.

 

(01°) let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists.

 

(02°) Then it is easy to show, A+B>C>A>B>U>0, where U=A+B-C.

 

Let us lead the number U (in the prime base n) to the form

(03°) u =Ud=(n^k)(n^p-1) with the aid of the multiplication of the equality 01° by the appropriate number d^n (which, as is known, exists), as a result of which the equality of 01° is converted into the equality

(1°) a^n+b^n-c^n = 0,

(2°) where a+b>c>a>b>u>0,

(3°) u=a+b-c, where the number u has r digits (but the highest digit of the number c let us designate by letter p) and, let us note, that

(4°) the number u is EVEN.

 

Let us try to find the digital solution of equation (3°) lower the r-digit endings are isolated from the high-range digits by two skew lines. Let

(5°)

a*=…?v//x…,

+

b*=…?v//y…,

-

c*=…?w//z…

==========

u = …00//9…

 

It is easy to see that the sum of the numbers

1. x+y is as the minimum equal to 9

2. Consequently, x is as the minimum equal to 9/2.

3. Consequently, z is as the minimum equal to 9/2.

4. Consequently, x+y is as the minimum equal to 9+9/2…

And so on to the eventual result:

5. x=y=z=9.

6. Consequently, v is as the minimum equal to 1.

However, on this the calculation of numbers to end does not can, since in this case u_(r+1) is not equal to zero. In order to remove this contradiction, is necessary number w_ (r+1) to make equal to one.

Of course other numbers, except these two, can enter into the discharges of higher than the r-th; however, when they satisfy the requirements: their sum is even.

 

But then the sum of all numbers, which figure in the Fermat’s equation, will be ODD. Consequently, and the number of a^n+b^n-c^n will be ODD, which is impossible

 

Thus, FLT is proven completely and indisputably.

 

Alas, the idea of proof proved to be hopeless. Long time out.

[Victor Sorok]

Equality-double in FLT. (Passed moment of the rejected idea)

 

Beginning is identical:

 

Designations:

a_(i) – the i-th digit from the end in the number a in the prime base n> 2.

a_(p, r) – the number, comprised of the digits from p to r of the number a.

9 – designation of digit n-1,

8 – designation of digit n-2.

 

Proof of FLT

 

(01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 in the integers exists.

(02°) Then, as it is easy to show, A+B>C>A>B>U>0, where U=A+B-C.

 

Let us lead (in prime base n) the number u=a+b-c to the form:

(03°) u =Ud= [n^(p-r)-1)](n^r)

with the aid of the multiplication of the equality of 01° by the corresponding number d^n (which, as is known, in the prime base n exists), as a result of which the equality of 01° is converted into the equality

(1°) a^n+b^n-c^n = 0, where a+b>c>a>b>u>0 and u=a+b-c.

 

Let us break all digits of the number u into three intervals:

1) [p, q+1],

2) [q, r+1],

3) [r, 1], where

p is greatest rank of the number c,

q is the greatest rank of the significant part of the number u,

r is greatest rank of the zero ending of the number u.

I.e. the number

(04°) u = u_(p, q+1)n^q+u_(q, r+1)n^r+u_(r, 1) = /00… 00//99… 99//00… 00/.

 

 

 

Let us examine first the case a_(R, 1)+b_(R, 1)-c_(R, 1)=0

 

Is attained the following process:

 

Let us replace all digits in the numbers a, b, c before (i.e. it is less) the range p by zero. We will obtain the numbers a*_(p, p), b*_(p, p), c*_(p, p)

(here it is below for simplicity of formulas zero endings equal for all numbers we do not write! Either it is convertible all numbers into n-mal fractions with the integer parts of a_p, b_p, c_p)

…and value of function F (a*, b*, C*), or F_ (p).

 

Then let us enter in the numbers a*, b*, c* the digits of the previous (i.e. the smaller) range p -1 and let us calculate value F_ (p, p -1) with the values of the numbers a_(p, p -1), b_(p, p -1), c_(p, p -1).

 

Then let us enter in the numbers a*, b*, c* the digits of range p -2 let us calculate value F_(p, p-2) with the values of the numbers a_(p, p-2), b_(p, p-2), c_(p, p-2).

 

And so on - up to the restoration of all digits of the numbers a, b, c.

 

***

Taking into account the formula of 4°, it is easy to see that the value

(5°) F_(p, q+1)= a_(p, q+1)^n+b_(p, q+1)^n-c_(p, q+1)^n<0.

 

(6°) In the interval from (p, q) to (p, r+1) function F(a*, b*, C*) it will be strictly increasing.

 

(7°) A in the interval from (p, r+1) to (p, 1) function F (a*, b*, C*) it will be as a whole of that diminishing.

 

***

Thus, function F (a*, b*, C*) in the closed interval [p, 1] in the first subinterval appears as a whole of that diminishing, in the second subinterval - strictly increasing, on the third - as a whole of that diminishing.

 

But the hence it follows that extrapolated diagram of the function F (a*, b*, C*) from (a*_(p, p), b*_(p, p), c*_ (p, p)) to (a, b, c) has also THE FRACTIONAL solution (a' , b' , c'), moreover 0<a'<a,0<b'<b, 0<c'<c!!!

 

There is above how to think over!

 

Continuation follows.

 

++++++++++++++++

 

Interpretation and the conclusion:

 

Either the numbers a, b, c are not given to the canonical form

or (a^n)_(1) = -a_(1), (b^n)_(1) = -b_(1), (c^n)_(1) = -c_(1),

that contradicts Little fermat's theorem.

 

Interpretation and the conclusion:

....

 

“The zero shot”: n=2

 

In this case

(- a^n) _ (1) = a_ (1)=1, (- b^n) _ (1) = b_ (1)=1, (- c^n) _ (1) = c_ (1)=1,

and there is no contradiction.

Now it is possible to study the formulation of text.

 

Simplified version of the proof

 

Let us assume that the equation

(1°) x^n+y^n-z^n = 0 has the entire solution (x, y, z) in the prime base n.

 

Let us designate the last digits of the numbers x, y, z, through a, b, c and the number x, y, z, in which the numbers a, b, c are substituted with zero, through A, B, C, i.e.,

(2°) A=x-a, B=y-b, C=z-c.

 

Proof of FLT

 

Let us rewrite equality (1°) in the form:

(3°) (A+a)^n+ (B+b)^n- (C+c)^n = 0, or

(4°) (A^n+A') + (B^n +B')- (C^n +C') = 0, or

(5°) (A^n+B^n-C^n) + (A' +B' - C') = 0, where:

(6°) A^n+B^n-С^n>0 (<0), if a+b-c=0 (if a+b-c=1),

(7°) the numbers A' , B' , C' finish respectively to the digits a, b, c and

(8°) the numbers A' +B' - C' < 0 (or >0), if a+b-c=0 (if a+b-c=1).

Consequently, or A'<0, or B'<0, or C'>0.

But then:

or in the number A^n +A' last digit is equal to …0-a=n-a

or in the number B^n +B' last digit is equal to …0-b=n-b

or in the number C^n +C' last digit is equal to …0-c=n-c.

Therefore, в одном из чисел A^n, B^n, С^n нарушается малая теорема Ферма.

So, допустив наличие решения уравнения 1°, мы получили противоречие.

 

Therefore, in one of the numbers A^n, B^n, С^n is disrupted Little Fermat's theorem.

So, after allowing the presence of the solution of the equation of 1°, we obtained contradiction.

 

FINISH

 

One additional reason:

 

Let the numbers x^n, y^n, z^n finish to the digits a, b, c.

Then:

From x^n+y^n-z^n= (2x^n+2y^n-2z^n) /2= [(x^n+y^n) - (z^n-y^n) - (z^n-x^n)]/2 and (!!!) the equality x^n+y^n-z^n=0 it follows that during the multiplication of the expression of x^n+y^n-z^n by positive number 2 the last digits a, b, c in the numbers of x^n, y^n, z^n are opposite to last digits -a, -b, -c in the identically equal to them numbers (x^n+y^n), (z^n-y^n), (z^n-x^n).

 

It is not strange whether?

[Victor Sorok]

FINISH

Proof of the FLT in the binary system of numeration for the prime n> 2.

 

Designations:

a_(i) – the i-th number from the end in the number a.

a_(p, q) – the number, comprised of the digits of the number a from the rank p to the rank q.

 

***

 

(01°) Let us assume the solution of the equation A^n+B^n-C^n = 0 0 in positive integer numbers exists.

(02°) While it is easy to show that A+B>C>A>B>U>0, where U=A+B-C.

 

Let us lead the number U (in the binary system) to the form

(03°) u =Ud=(2^r)(2^p-1) with the aid of the multiplication of the equality of 01° by the appropriate number d^n (which, as is known, exists),

as a result what the equality 01° is converted into the equality

(1°) a^n+b^n-c^n = 0 where, as is well known of the theory of the Fermat’s equality,

(2°) a+b>c>a>b>u>0, c>a>c(n-1)/n;

(3°) u=a+b-c=u_(q, r+1)(2^r);

(4°) c-a=b-u=g>0;

(5°) b/u<2;

(6°) (c-a)/a<1/n.

 

***

 

Let us break all ranks of the number u into three intervals:

1) [p, q+1],

2) [q, r+1],

3) [r, 1], where

p – is the greatest rank of the number c,

q – is the greatest rank of the significant part of the number u,

r – is the greatest rank of the zero end of the number u.

 

I.e. the number

 

(7°) u = u_(p, q+1)(2^q)+u_(q, r+1)(2^r)+u_(r, 1) = /00…00//11…11//00…00/,

 

and the equality a+b-c=u takes approximately this form:

 

a= /1…11//11…10//x…/,

+

b= /0…01//00…01//y…/,

-

c= /1…10//00…01//z…/.

==============

u=/0…00//11…11//00…00/.

 

Let us carry out the digital analysis of this equality (principal moment of the proof).

 

(8°) The number b_(p, q+1) can be equal only to 1 (which follows from 4° and 6°).

(9°) If the digit a_(q+1)=1, then follows from the equality c_(p, q+2)=a_(p, q+2) that a>c, that contradicts to 2°.

(10°) But if a_(q+1)=0, then c_(p, q+1)-a_(p, q+1)=1 (here 1 is a digit of the rank q+1). A difference b-u, as can be seen from a numerical example, is considerably less than 2^q. And in this case we have contradiction with 5°.

 

But since the third it is not given, equality 1° does not have a solution. FLT is proven.

 

***

 

P.S. However, the proof remains true and in the prime base n> 2. Thus, after small correction the previous proof is true.

[Victor Sorok]

Before shutting after itself door…

 

From the identity (with a+b-c=0)

(1°) (a+1)(b+1)(c+1)=abc+c^2+1 it follows that

(2°) (a+1)(b+1)(c+1)-c^2=abc+1.

 

In the Fermft’s equality the role of the numbers a, b, c play the numbers a^n, b^n, c^n.

 

After the multiplication of Fermft’s equality by the sufficiently large number

(3°) ones into 2° can be disregarded, and

(4°) 0<<ab-c<<ab.

 

And the equality 2° is converted into the inequality.

Edited August 15, 2008 by Victor Sorok

[Victor Sorokine]

In 1990 I proved FLT for the case when c is even.

Today I found proof, also, for even a /or b/. Here is it.

 

Designations:

p - an integer,

q - an odd number,

X - set of odd numbers of the type of x=2q+1,

Y - set of odd numbers of the type of y=4p+1.

Let us name two odd numbers x' and x'' uniform.

Let us name two odd numbers x and y diverse (different-type).

 

The following assertions are obvious:

(1°) if odd numbers a and b are of the same type, then the numbers a+2q and b are diverse.

(2°) the numbers of 2p+q and 2p- q are diverse.

 

Proof of FLT for the odd of n> 2.

 

Let us assume

(3°) a^n+b^n=c^n, or

(4°) (c-b)P+(c-a)Q=(a+b)R,

where two of the numbers a, b, c and of the numbers c-b, c-a, a+b are odd and

(5°) c>a>b>0.

 

Case 1. Number c is even.

Let us select the pair of the numbers of a+b=e (even) and of c-a=d (odd)

and let us examine the pair of numbers of d+e and d-e, which are been, obviously, DIVERSE (DIFFERENT-TYPE) (see 2°).

From the other side, THE SAME numbers of d+e (=c+b) and d-e (=c-b-2a) are UNIFORM, since the numbers of c+b and c-b are diverse, the numbers c-b and c-2a are also diverse, therefore, the number c+b and c-b-2a are uniform.

And we have a contradiction: the numbers in two identical pairs of numbers are different-type and at the same time uniform.

 

 

Case 2. Number a /or b/ even.

 

After the simplest substitution –

a=2^k-a*, where 2^k>c+a, –

the proof of this case for the numbers c, a*, b is completely analogous to previous proof.

 

(It is certain, - “this it cannot be according to the basic postulate of official science, because this it cannot be ever!” - my proof is not correct.)

 

Aug 26, 2008

[Victor Sorokine]

It seems that Pierre Fermat was rights…

 

Let us assume that for the integers a, b, c there is a equality

(1°) a^n+b^n=c^n, where simple n> 2.

 

Let us examine the equality of 1° in the prime base

(2°) q=pn+1>3c^ {2n} (it is known that the set of such numbers q is infinite).

 

It is easy to see that the number

(3°) D=c^{qn} - a^{qn} - b^{qn} is divided by q. But

(4°) D=c^{npn+n} - a^{npn+n} - b^{npn+n}, where the numbers

(5°) c^{np}, a^{np}, b^{np} finish to digit 1 and, therefore, the sum of the last digits of the number D into 4° will be equal to q:

(6°) c^{2n} - a^{2n} - b^{2n} =q, that contradicts 2°.

 

(Sep 13, 2008)

[Victor Sorokine]

It seems that Pierre Fermat was rights…

 

Interesting fact: in the Fermat’s equality

U=(a+b)-(c-b)-(c-a)<2u=2(a+b-c)!

 

Let for integers a, b, c and odd n>2 there is a equality

(1°) a^n+b^n-c^n, where the number

(2°) u=a+b-c>0.

 

Then in the equivalent equality

(3°) (a+b)R-(c-b)P-(c-a)Q=0, where the numbers P, Q, R are known polynomials of the decomposition of the sum of two degrees,

the number

(4°) U=(a+b)-(c-b)-(c-a)=2u.

However, this equality is not carried out.

 

Are actual, with equal P, Q, R - for example, with

(5°) P=Q=R=a^(n-1) –

the equality of 4° is correct.

However, in comparison with the equality of 5° in the equality of 4° coefficient R attached to positive item (a+b) IS LESS a^(n-1) and coefficients P and Q attached to negative items - (c-b) and (c-a) ARE MORE a^(n-1).

As a result the number

(6°) U=(a+b)-(c-b)-(c-a)<2u=a+b-c.

And we have a contradiction since (a+b)-(c-b)-(c-a)=2(a+b-c).

[Victor Sorokine]

It seems that Pierre Fermat was rights…

 

It seems, we arrived!

 

Here is actually the complete text of the proof of the FLT.

 

In the prime base n>2 in the equality

(1°) a^n+b^n-c^n=0

the number (sum of digits)

(2°) a+b-c=U=un^k, where k>0 and u not it is divided by n.

But then

(3°) U+0=(a+b-c)+a^n+b^n-c^n=a(1+a^{n -1}) +b(1+b^{n -1})-c(1+c^{n -1})=U.

 

But the equality 3° is contradictorily, since if abc not is divided by n, then the last non-zero digit in the number U has simultaneously TWO different values: in the number U+0 it is doubly (since, taking into account to Fermat's litle theorem, all three expressions in the parentheses finish are finished by digit 2) more than in the number U (but if the doubled value exceeds n, then from it should be deducted base n).

 

But if, for example, the number of c=c'n^t, where c' not is multiple n and t>0, then in the new value of the number U [t (n -1) of +1]-th (from the end) digit in the number c will increase by 1, and tn-th digit in the number a+b will be doubled.

 

Here, strictly, and everything.

 

2008.10.10

 

But this is how appears brief proof of the FLT (for the basic case: abc not divided by ghbme n> 2):

 

In the Fermat’s equality

the number u=a+b-c IS NOT EQUAL to the number

U+0=(a+b-c)+a^n+b^n-c^n=a(1+a^{n-1})+b(1+b^{n-1})-c(1+c^{n-1}),

since in the second number all expressions in the parentheses finish, according to Fermat's Little theorem by digit 2 and, therefore, the last significant (non-zero) digit of the number U+0 is twice more than as the last significant digit in the number U.

 

In order to refute my proof, it is necessary to prove that:

The multiplication of single-digit positive number by 2 does not change its value.

 

But to whom this on the teeth?