Fermat's last theorem is: contradiction on the parity

[Victor Sorok]

Complet text of the proof

 

 

All wholes numbers in the next are given in the prime base [math]n[/math].

 

Here are the known positions, repeatedly discussed on the mathematical forums. Thus, let

 

(1°) [math]A^n+B^n = C^n[/math], where prime [math]n > 2[/math] and [math]A, B, C[/math] have no common factors.

Let for the moment the number [math]ABC[/math] is not divided by [math]n[/math]. Then:

 

(2°) [math]C^n-B^n=(C-B)P[/math], where [math]C-B=a'^n, P=a''^n[/math],

[math]C^n-A^n=(C-A)Q[/math], where [math]C-A=b'^n, Q=b''^n[/math],

[math]A^n+B^n=(A+B)R[/math], where [math]A+B=c'^n, R=c''^n[/math],

[this follows from the simple lemma: if the numbers [math]A[/math] and [math]B[/math] have no common factors and the number [math]A+B[/math] is not divided by [math]n[/math], then the numbers [math]A+B[/math] and [math]R = \frac{A^n+B^n}{A+B}[/math] have no common factors (since [math]R=(A+B)^2T+nAB)[/math]];

 

(3°) the numbers [math]a', a'', b', b'', c', c''[/math] have no common factors;

 

(4°) [math]A=a'a'', B=b'b'', C=c'c''[/math].

 

(5°) Important equalities:

[math]c'^n - b'^n = 2a'a'' - (C-B)[/math],

[math]c'^n - a'^n = 2b'b'' - (C-A)[/math],

[math]a'^n + b'^n = 2c'c'' - (A+B)[/math], or

 

(5°a) [math](A+B) - b'^n = 2a'a'' - a'^n [/math],

[math](A+B)- a'^n = 2b'b'' - b'^n [/math],

[math]a'^n + b'^n = 2c'c'' - (A+B)[/math],

from here we find the numbers [math]a''[/math] and [math]b''[/math]:

 

(6°) [math]a'' = \frac{(A+B) - b'^n + a'^n}{2a'}[/math],

[math]b'' = \frac{(A+B) - a'^n + b'^n}{2b'}[/math].

At last, we find two important formulas, лежащие in the base of the proof of the FLT:

(7a°) [math]a'' - b'' = \frac{(b'-a')(A+B) + (a'+b')(a'^n-b'^n)}{2a'b'}[/math],

(7b°) [math]a'' + b'' = \frac{(b'+a')(A+B) - (a'-b')(a'^n-b'^n)}{2a'b'}[/math].

 

Elementary proof of the Fermat's last theorem

 

Let for distinctness the numbers [math]A[/math] and [math]B[/math] is not divided by [math]n[/math]. It is easy to see that the formulas 7° remain correct and if the number [math]C[/math] is not divided by [math]n[/math].

Let us examine the last two-three significant digits in the numbers [math]a'', b'', P, Q[/math] in binary system. Here are still the possible cases:

 

a) if [math]a'=[/math]…[math]01[/math], [/math]b'=[/math]…[math]01[/math] [or [math]a'=[/math]…[math]11, b'=[/math]…[math]11[/math]], then [math]a''+ b''=[/math]…[math]00[/math], but [math]P+Q=[/math]…[math]10[/math];

 

b) if the number [math]a'+b'[/math] has [math]k[/math] zeros on the end, then the number [math]a''-b''[/math] has [math]k[/math] zeros on the end, but the number [math]P-Q[/math] has [math]k+1[/math] zeros on the end [since the number [math]P-Q[/math] is divided by [math]a-b[/math] and, moreover, by common divisor of the numbers [math]c[/math] and [math]a+b[/math]];

 

c) if [math]a'=[/math]…[math]0[/math] [or [math]b'=[/math]…[math]0[/math]], then both numbers [math]a''- b''[/math] and [math]a''+ b''[/math] are divided by [math]4[/math], that is impossible since the numbers [math]a''[/math] and [math]b''[/math] have no common factors.

 

If the number [math]A[/math] is divided by [math]n[/math], then they are taken the numbers [math]c'', b'', R, Q[/math]. Their analysis leads to the analogous contradictions:

 

(8a°) [math]c'' - b'' = \frac{(c'-a')(C+B) - (c'+b')(C-B)}{2a'b'}[/math],

(8b°) [math]c'' + b'' = \frac{(c'+a')(C+B) - (c'-b')( C-B)}{2a'b'}[/math].

 

Takeing into account school formulas for polynomials [math]R[/math] and [math]Q[/math], , it is easy to see that the number [math]R - Q[/math] is divided by even number [math]C+B[/math] and, furthermore, to the common divisor even numbers [math]A[/math] and [math]C-B[/math]. Consequently, number [math]C+B[/math] is divided by 2 only to the first power.

 

Let us examine the last two-three significant digits in the numbers [math] c'', b'', R, Q [/math] in binary system. Here are still the possible cases:

a) if [math]c'=[/math]…[math]01, b'=[/math]…[math]01[/math] [or [math]c'=[/math]…[math]11, b'=[/math]…[math]11[/math]], then [math]c''+ b''=[/math]…[math]00[/math], but [math]R+Q=[/math]…[math]10[/math];

b) if the number [math]c'+b'[/math] has [math]k[/math] zeros on the end, then the number [math]c''-b''[/math] has [math]k[/math] zeros on the end, but the number [math]R-Q[/math] has [math]k+1[/math] zeros on the end [since the number [math]R-Q[/math] ] is divided by [math]C-B[/math] and, moreover, by common divisor of the numbers [math]A[/math] and [math]C+B[/math]];

c) if [math]c'=[/math]…[math]0[/math] [or [math]b'=[/math]…[math]0[/math]], then both numbers [math]c''- b''[/math] and [math]c''+ b''[/math] are divided by [math]4[/math], that is impossible since the numbers [math]c''[/math] and [math]b''[/math] have no common factors.

 

(It is necessary to keep in mind that the number [math]R[/math] [is analogous and [math]P, Q[/math]] is even only if both number [math]A[/math] and [math]B[/math] are even.)Thus, FLT is completely proven.

 

Victor Sorokine

 

Mezos, 2 August 2007

[Victor Sorok]

Mini-proof of the case [MATH]a'+b'[/MATH] is divided exact by two:

Then numbers [MATH]a'- b'[/MATH], therefore, and [MATH]a'^n-b'^n[/MATH] or [MATH]A-B[/MATH], therefore and [MATH]A^n-B^n[/MATH], are divided by or [MATH]4[/MATH]. But sum and difference of odd numbers [MATH]A^n[/MATH] and [MATH]B^n[/MATH] cannot simultaneously be divided into [MATH]4[/MATH]. And contradiction is present.

[Victor Sorok]

Turning (after MINI-proof) in the of the FLT. Complete text from the very beginning...

 

Thus, let

(1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and (for the beginning) [math]a[/math] and [math]b[/math] are odd and [math]c=2^kx[/math], where here and here and everywhere further the number [math]x[/math] is odd, value of which DOES NOT HAVE an importance.

It is easy to see that

(2°) [math]a+b-c=2^kx[/math].

 

Let us show that equality 1° in the wholes numbers is insoluble.

 

Let us make the first substitution:

(3-1°) [math]a=c_1-b_1, b=c_1-a_1, c=a_1+b_1[/math], where

(4-1°) [math]a_1+b_1-c_1=2^{k-1}x[/math].

It is easy to see that

(5-1°) [math]a_1+b_1=2^kx[/math] and the numbers [math]a_1, b_1, c_1[/math] are whole, but the number [math]a_1[/math] and [math]b_1[/math] are odd.

 

Let us make the second substitution:

(3-2°) [math]a_1=c_2-b_2, b_1=c_2-a_2, c_1=a_2+b_2[/math], where

(4-2°) [math]a_2+b_2-c_2=2^{k-2}x[/math].

It is easy to see that

(5-2°) [math]a_2+b_2=2^{k-1}x[/math] and the numbers[math]a_2, b_2, c_2[/math] are whole, but the numbers [math]a_2[/math] and [math]b_2[/math] are odd.

……………………………………….

And so on to [math]k[/math]-th substitution:

(3-k°) [math]a_{k- 1}=c_k-b_k, b_{k-1}=c_k-a_k, c_{k- 1}1=a_k+b_k[/math], where

(4-k°) [math]a_k+b_k-c_k=2^kx[/math].

It is easy to see that

(5-k°) [math]a_k+b_k=2^1x[/math] and the numbers [math]a_k, b_k, c_k[/math] are whole, but the numbers [math]a_k[/math] and [math]b_k[/math] are odd.

But then from 4-k° we have:

(6°) [math]a_k-b_k=2^2y[/math], where [math]y[/math] is whole number, value of which DOES NOT HAVE an importance.

 

Now let us express the numbers [math]a, b, c[/math] через [math]a_k, b_k, c_k [/math].

At that equal items in the numbers [math]a[/math] and [math]b[/math] we will reject, since to us is important only the difference of the numbers [math]a[/math] and [math]b[/math]. As a result, obviously, we obtain the equalities:

(7°) [math]a=b_k+x[/math] and [math]b=a_k+x[/math] (or [math]a=a_k+x[/math] and [math]b=b_k+x[/math]), where [math]x[/math] – rejected the number.

And now, after calculating the number [math]a^n-b^n[/math], we find that it IS MULTIPLE FOUR. But, as can be seen from 1°, and the number [math]a^n-b^n[/math] is also MULTIPLE FOUR, that by odd numbers [math]a^n[/math] and [math]b^n[/math] is impossible.

 

FLT is proved.

 

Turning (after MINI-proof) in the of the FLT. Complete text from the very beginning...

 

Thus, let

(1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and (for the beginning) [math]a[/math] and [math]b[/math] are odd and [math]c=2^kx[/math], where here and here and everywhere further the number [math]x[/math] is odd, value of which DOES NOT HAVE an importance.

It is easy to see that

(2°) [math]a+b-c=2^kx[/math].

 

Let us show that equality 1° in the wholes numbers is insoluble.

 

Let us make the first substitution:

(3-1°) [math]a=c_1-b_1, b=c_1-a_1, c=a_1+b_1[/math], where

(4-1°) [math]a_1+b_1-c_1=2^{k-1}x[/math].

It is easy to see that

(5-1°) [math]a_1+b_1=2^kx[/math] and the numbers [math]a_1, b_1, c_1[/math] are whole, but the number [math]a_1[/math] and [math]b_1[/math] are odd.

 

Let us make the second substitution:

(3-2°) [math]a_1=c_2-b_2, b_1=c_2-a_2, c_1=a_2+b_2[/math], where

(4-2°) [math]a_2+b_2-c_2=2^{k-2}x[/math].

It is easy to see that

(5-2°) [math]a_2+b_2=2^{k-1}x[/math] and the numbers[math]a_2, b_2, c_2[/math] are whole, but the numbers [math]a_2[/math] and [math]b_2[/math] are odd.

……………………………………….

And so on to [math]k[/math]-th substitution:

(3-k°) [math]a_{k- 1}=c_k-b_k, b_{k-1}=c_k-a_k, c_{k- 1}=a_k+b_k[/math], where

(4-k°) [math]a_k+b_k-c_k=2^1x[/math].

It is easy to see that

(5-k°) [math]a_k+b_k=2^1x[/math] and the numbers [math]a_k, b_k, c_k[/math] are whole, but the numbers [math]a_k[/math] and [math]b_k[/math] are odd.

But then from 4-k° we have:

(6°) [math]a_k-b_k=2^2y[/math], where [math]y[/math] is whole number, value of which DOES NOT HAVE an importance.

 

Now let us express the numbers [math]a, b, c[/math] через [math]a_k, b_k, c_k [/math].

At that equal items in the numbers [math]a[/math] and [math]b[/math] we will reject, since to us is important only the difference of the numbers [math]a[/math] and [math]b[/math]. As a result, obviously, we obtain the equalities:

(7°) [math]a=b_k+x[/math] and [math]b=a_k+x[/math] (or [math]a=a_k+x[/math] and [math]b=b_k+x[/math]), where [math]x[/math] – rejected the number.

And now, after calculating the number [math]a^n-b^n[/math], we find that it IS MULTIPLE FOUR. But, as can be seen from 1°, and the number [math]a^n-b^n[/math] is also MULTIPLE FOUR, that by odd numbers [math]a^n[/math] and [math]b^n[/math] is impossible.

 

FLT is proved.

[Moshe_Klein]

Shalom Victor,

 

I joint today to this Forum - for your FLT paper.

I will read your post again and respond soon.

Thank you for sharing your work.

Moshe:eyebrow:

[Victor Sorok]

Elucidating trifles

 

1. Formulation of the contradiction:

Sum and difference in the odd numbers cannot simultaneously be divided into 4.

 

2. Details about the [math]i[/math]-th substitution:

 

(3-i°) [math]a_{i-1}=c_i-b_i, b_{i-1}=c_i-a_i, c_{i- 1}=a_i+b_i[/math], where

(4-i°) [math]a_i+b_i-c_i=2^{k-i-1}x[/math].

It is easy to see that

(5-i°) [math]a_i+b_i=2^{k-i}x[/math], [math]c_i=2^{k-i-1}x[/math], and the numbers [math]a_i, b_i, c_i[/math] are whole, but the numbers [math]a_i[/math] and [math]b_i[/math] are odd.

Thus, after the latter ([math]k[/math]-th) substitution all numbers [math]a_k, b_k, c_k[/math] become odd.

 

3. The case with even [math]a[/math] proves in perfect analogy, but with the only difference that equal in the absolute value items [math]y[/math] in the numbers [math]c_k[/math] and [math]b_k[/math] have the opposite signs.

And now after the substitution of the values [math]c[/math] and [math]b[/math], expressed through the numbers [math]c_k[/math] and [math]b_k[/math], we obtain the contradiction: both numbers [math]c^n-b^n[/math] and [math] c^n+b^n [/math] are divided by [math]4[/math], that - if [math]c^n[/math] and [math]b^n[/math] are odd – it is impossible.

 

Elucidating trifles

 

1. Formulation of the contradiction:

Sum and difference in the odd numbers cannot simultaneously be divided into 4.

 

2. Details about the [math]i[/math]-th substitution:

 

(3-i°) [math]a_{i-1}=c_i-b_i, b_{i-1}=c_i-a_i, c_{i- 1}=a_i+b_i[/math], where

(4-i°) [math]a_i+b_i-c_i=2^{k-i-1}x[/math].

It is easy to see that

(5-i°) [math]a_i+b_i=2^{k-i}x[/math], [math]c_i=2^{k-i-1}x[/math], and the numbers [math]a_i, b_i, c_i[/math] are whole, but the numbers [math]a_i[/math] and [math]b_i[/math] are odd.

Thus, after the latter ([math]k[/math]-th) substitution all numbers [math]a_k, b_k, c_k[/math] become odd.

 

3. The case with even [math]a[/math] proves in perfect analogy, but with the only difference that equal in the absolute value items [math]y[/math] in the numbers [math]c_k[/math] and [math]b_k[/math] have the opposite signs.

And now after the substitution of the values [math]c[/math] and [math]b[/math], expressed through the numbers [math]c_k[/math] and [math]b_k[/math], we obtain the contradiction: both numbers [math]c^n-b^n[/math] and [math] c^n+b^n [/math] are divided by [math]4[/math], that - if [math]c^n[/math] and [math]b^n[/math] are odd – it is impossible.

 

Corr:

(5-i°) [math]a_i+b_i=2^{k-i+1}x[/math], [math]c_i=2^{k-i}x[/math], and the numbers [math]a_i, b_i, c_i[/math] are whole, but the numbers [math]a_i[/math] and [math]b_i[/math] are odd.

 

Elucidating trifles

 

1. Formulation of the contradiction:

Sum and difference in the odd numbers cannot simultaneously be divided into 4.

 

2. Details about the [math]i[/math]-th substitution:

 

(3-i°) [math]a_{i-1}=c_i-b_i, b_{i-1}=c_i-a_i, c_{i- 1}=a_i+b_i[/math], where

(4-i°) [math]a_i+b_i-c_i=2^{k-i-1}x[/math].

It is easy to see that

(5-i°) [math]a_i+b_i=2^{k-i+1}x[/math], [math]c_i=2^{k-i}x[/math], and the numbers [math]a_i, b_i, c_i[/math] are whole, but the numbers [math]a_i[/math] and [math]b_i[/math] are odd.

Thus, after the latter ([math]k[/math]-th) substitution all numbers [math]a_k, b_k, c_k[/math] become odd.

 

3. The case with even [math]a[/math] proves in perfect analogy, but with the only difference that equal in the absolute value items [math]y[/math] in the numbers [math]c_k[/math] and [math]b_k[/math] have the opposite signs.

And now after the substitution of the values [math]c[/math] and [math]b[/math], expressed through the numbers [math]c_k[/math] and [math]b_k[/math], we obtain the contradiction: both numbers [math]c^n-b^n[/math] and [math] c^n+b^n [/math] are divided by [math]4[/math], that - if [math]c^n[/math] and [math]b^n[/math] are odd – it is impossible.

 

Turning (after MINI-proof) in the of the FLT. Complete text from the very beginning...

 

Thus, let

(1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and (for the beginning) [math]a[/math] and [math]b[/math] are odd and [math]c=2^kx[/math], where here and here and everywhere further the number [math]x[/math] is odd, value of which DOES NOT HAVE an importance.

It is easy to see that

(2°) [math]a+b-c=2^kx[/math].

 

Let us show that equality 1° in the wholes numbers is insoluble.

 

Let us make the first substitution:

(3-1°) [math]a=c_1-b_1, b=c_1-a_1, c=a_1+b_1[/math], where

(4-1°) [math]a_1+b_1-c_1=2^{k-1}x[/math].

It is easy to see that

(5-1°) [math]a_1+b_1=2^kx[/math] and the numbers [math]a_1, b_1, c_1[/math] are whole, but the number [math]a_1[/math] and [math]b_1[/math] are odd.

 

Let us make the second substitution:

(3-2°) [math]a_1=c_2-b_2, b_1=c_2-a_2, c_1=a_2+b_2[/math], where

(4-2°) [math]a_2+b_2-c_2=2^{k-2}x[/math].

It is easy to see that

(5-2°) [math]a_2+b_2=2^{k-1}x[/math] and the numbers[math]a_2, b_2, c_2[/math] are whole, but the numbers [math]a_2[/math] and [math]b_2[/math] are odd.

……………………………………….

And so on to [math]k[/math]-th substitution:

(3-k°) [math]a_{k- 1}=c_k-b_k, b_{k-1}=c_k-a_k, c_{k- 1}1=a_k+b_k[/math], where

(4-k°) [math]a_k+b_k-c_k=2^0x[/math].

It is easy to see that

(5-k°) [math]a_k+b_k=2^1x[/math] and the numbers [math]a_k, b_k, c_k[/math] are whole, but the numbers [math]a_k[/math] and [math]b_k[/math] are odd.

But then from 4-k° we have:

(6°) [math]a_k-b_k=2^2y[/math], where [math]y[/math] is whole number, value of which DOES NOT HAVE an importance.

 

Now let us express the numbers [math]a, b, c[/math] через [math]a_k, b_k, c_k [/math].

At that equal items in the numbers [math]a[/math] and [math]b[/math] we will reject, since to us is important only the difference of the numbers [math]a[/math] and [math]b[/math]. As a result, obviously, we obtain the equalities:

(7°) [math]a=b_k+x[/math] and [math]b=a_k+x[/math] (or [math]a=a_k+x[/math] and [math]b=b_k+x[/math]), where [math]x[/math] – rejected the number.

And now, after calculating the number [math]a^n-b^n[/math], we find that it IS MULTIPLE FOUR. But, as can be seen from 1°, and the number [math]a^n-b^n[/math] is also MULTIPLE FOUR, that by odd numbers [math]a^n[/math] and [math]b^n[/math] is impossible.

 

FLT is proved.

[Moshe_Klein]

Victor , why It is easy to see that

(2°) ... a+b-c= 2^k * x ?

 

Moshe

[Victor Sorok]

Victor , why It is easy to see that

(2°) ... a+b-c= 2^k * x ?

 

Moshe

Since a+b= 2^kn * x and c= 2^k * x

BUT!

 

1. Alas, and this proof proved to be incorrect in PRINCIPLE.

2. However, if elementary proof FLT exists, then it is located somewhere entirely next. I analyze errors and new properties of Fermat’s equality. Sometimes I will present the logic of this analysis, and the thus far basic conclusion: contradiction must be revealed only with the aid of THE INEQUALITIES.

3. Long ago I advanced this hypothesis:

since the number [math]P-Q[/math] is divided by [math]A-B[/math] and, moreover, by common divisor of the numbers [math]c[/math] and [math]a+b[/math], then

in number [math]P-Q= a''^n -b''^n[/math] precisely cofactor [math]a''- b"[/math] is divided by [math] (A-B)c'[/math].

But prove this for me everything could not.

4. On 2 August I proposed proof with the aid of the number

[math]a''-b'' = \frac{(b'-a')(A+B) + (a'+b')(a'^n-b'^n)}{2a'b'}[/math] (cf. formula (7a°) #1 on 08-11-2007, 09:39 PM).

And here only today I noted that number [math]a''- b''[/math] is divided also by [math] (A-B)c'[/math].

5. Now it is possible to show that the number [math]\frac{ a''-b''}{(A-B)c'}[/math] is IMPROPER FRACTION, that contradicts p.3.

(Thus, to me it is obvious that [math]a''-b''< (\frac{P-Q}{a''-b''})^{\frac{1}{n}}< c'[/math] etc.)

 

Thus, I gradually approach the simple calculations.

 

Splendid confirmation of the hypothesis!

 

Calculation of the difficultest case:

[math]n=3[/math], the numbers [math]A and B[/math] are not divided by [math]n[/math].

 

If [math]A=100[/math] and [math]B=200[/math], then

[math]a''-b''=7,9834[/math] and[math](a'-b')c' = 18,476[/math].

Therefore the fraction [math]\frac{a''-b''}{(a'-b')c'}[/math] is NOT whole!

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

Calculations:

[math]a=100[/math], [math]b=200[/math], [math]c=208,008[/math];

[math]a^3=1000000[/math], [math]b^3=8000000[/math], [math]c^3=9000000[/math];

[math]c-b=8,008[/math], [math]c-a=108,008[/math], [math]a+b=300[/math];

[math]a_1=2,001[/math], [math]b_1=4,762[/math], [math]c_1=6,694[/math];

[math]a_2=49,983[/math], [math]b_2=41,999[/math], [math]c_2=31,074[/math];

[math]b_2-c_2=10,925[/math], [math]a_2-c_2=18,909[/math], [math]a_2-b_2=7,983[/math];

[math]c_1+b_1=11,456[/math], [math]c_1+a_1=8,695[/math], [math]b_1-a_1=2,761[/math];

[math] (c_1+b_1)a_1[/math]: [math](c_1+a_1)b_1[/math]: [math] (b_1-a_1)c_1[/math]:

[math] 22,923[/math]; [math] 41,406[/math]; [math]18,482[/math].

 

Splendid confirmation of the hypothesis!

 

Another confirmation of the hypothesis:

 

If [math]n=3[/math], [math]a=180[/math] и [math]b=200[/math], then we have:

 

[math]a_2-b_2=11,79[/math], but [math] (b_1-a_1)4n=14,844[/math] (i.e. 11,79<14,844),

[math]c_2+b_2=37,287[/math], but [math] (c_1+a_1)4n=134,388[/math] (i.e. 37,287<134,388),

[math]ca_2+a_2=25,497[/math], but [math] (c_1+a_1)4n=149,232[/math] (i.e. 25,497<149,232).

 

+++

 

It is obvious, when [math]a[/math] approach to [math]b[/math], then

[math]f = (b_1-a_1)4n[/math] approach to [math]g = a_2-b_2[/math].

Apropos, from formula 7° it follows that [math]g = a_2 - b_2[/math] is divided by [math]4n(a_1 - b_1)[/math], but – as calculations show in concrete numbers - (8°) [math]g = a_2 - b_2 < 4n(a_1 - b_1)[/math]. It is the CONTRADICTION of the Fermat’s equality.

[Moshe_Klein]

Dear Victor,

 

Ok, there was a mistake in the prove.

But what is the bottom line right now ,

Have you prove FLT or not ?

 

Moshe

 

[Victor Sorok]

Dear Victor,

 

Ok, there was a mistake in the prove.

But what is the bottom line right now ,

Have you prove FLT or not ?

 

Moshe

 

 

I had proved that

If a solution of the Fermat’s equality in whole numbers exists, then the number [math]g = a_2-b_2[/math] is divided by [math]4n(a_1 - b_1)[/math],

i.e. the number [math]h= \frac{a_2-b_2}{4n(b_1-a_1)}[/math] is whole.

On the other hand, calculations on computer for real number [math]a, b, c[/math], which answer to the Fermat’s equality, show that [math]0 < h << 1[/math].

Question: is it possible to show this fact?

V.S.

[Victor Sorok]

Dear Victor,

 

Ok, there was a mistake in the prove.

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

Deviation in the proof after the formula 7°.

 

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

I remind basic properties of the Fermat’s equality

 

(1°) [math]a^n+b^n = c^n[/math], where prime [math]n > 2[/math] and [math]a, b, c[/math] have no common factors.

Let for the time being the number [math]abc[/math] is not divided by [math]n[/math]. Then:

(2°) [math]c^n-b^n=(c-b)P[/math], where [math]c-b=a_1^n, P=a_2^n[/math],

[math]c^n-a^n=(c-a)Q[/math], where [math]c-a=b_1^n, Q=b_2^n[/math],

[math]a^n+b^n=(a+b)R[/math], where [math]a+b=c_1^n, R=c_2^n[/math],

[this follows from the simple lemma: if the numbers [math]A[/math] and [math]B[/math] have no common factors and the number [math]A+B[/math] is not divided by [math]n[/math], then the numbers [math]A+B[/math] and [math]R = \frac{A^n+B^n}{A+B}[/math] have no common factors (since [math]R=(A+B)^2T+nAB)[/math]];

(3°) the numbers [math]a_1, a_2, b_1, b_2, c_1, c_2[/math] have no common factors,

the numbers [math]P, Q, R[/math] and [math]a_2, b_2, c_2[/math] нечетны;

(4°) [math]a=a_1a_2, b=b_1b_2, b=c_1c_2[/math].

(5°) From analysis of difference of the polynomials [math]P[/math] and [math]Q[/math] in its standard (school) form one can see, that the number [math]P-Q[/math] is divided by [math]a-b[/math].

(6°) Important equalities:

[math]c_1^n - b_1^n = 2a_1a_2 - a_1^n= 2a - (c-b)[/math],

[math]c_1^n - a_1^n = 2b_1b_2 - b_1^n = 2b - (c-a)[/math],

[math]a_1^n + b_1^n = 2c_1c_2 - c_1^ n =2c - (a+b)[/math],

from which we find the base formula of the number [math]g=a_2 - b_2[/math]:

(7°) [math]g= a_2 - b_2 = \frac{(a+b)(b_1-a_1) + (a_1+b_1)( a_1^n-b_1^n)}{2a_1b_1}[/math], from which one can see, that the number [math]g=a_2-b_2[/math] is divided by [math]a_1-b_1[/math].

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

Next we write down the numbers [math]P-Q[/math] and [math]a-b[/math] in the form:

(8°) [math]P-Q= a_2^n-b_2^n=(a_2-b_2)W[/math] and

(9°) [math]a-b= a_1^n-b_1^n=(a_1-b_1)V[/math].

Now, comparing its efficient in (5°) and (7°), one can make interesting conclusion:

the number [math] a_2^n-b_2^n=(a_2-b_2)W[/math] is divided by [math]a_1^n-b_1^n(a_1-b_1)V [/math],

but the number [math]a_2-b_2[/math] is divided only by [math]a_1-b_1[/math].

Consequently, the number [math]W[/math] is divided by [math]V[/math].

 

BUT: if the numbers [math]a_1, b_1, a_2, b_2[/math] have no common factors (this follows from 2°) the numbers [math]W[/math] and [math]V[/math] have NO common factors.

 

(It is not inconceivable that this assertion constitute well-known theorem from the theory of discontinuous numbers. And if this lemma-theorem is known in mathematics, then elementary proff of FLT is on hand.)

 

Numerical example to lemma-theorem.

Let [math]n=3, a=1, b=4, c=2, d=3, a+b=5, c+d=5[/math], [math]a+b[/math] is divided by [math]c+d[/math], but [math]\frac{a^n+b^n}{a+b}=13[/math] is NOT divided by [math]\frac{c^n+c^n}{c+c}=7[/math].

 

Victor Sorokine. Mezos (France)

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

 

It is not inconceivable that this assertion constitute well-known theorem from the theory of discontinuous numbers. And if this lemma-theorem is known in mathematics, then elementary proff of FLT is on hand.

 

And since it is evident from the previous post, difficulty of FLT is a difficulty of final lemma-theorem.

Possibly, someone will give reference to its publication.

Possibly, someone will be able to give its (judging by the analogous theorems) simple proof.

Well and no - it is necessary to search for proof. The first (previous) numerical example illustrates the essence of theorem.

But here is more strongly example for the [math]a, b, c, d[/math], which have no common factors.

 

Let [math]c+d[/math] is divided by [math]a^n+b^n[/math].

Then [math]c+d[/math] is divided by [math]a+b[/math].

But [math]\frac{c^n+d^n}{c+d}[/math] and [math]c+d[/math] have no common factors (cf. notice to 2°). Therefore, [math]\frac{c^n+d^n}{c+d}[/math] is NOT divided by [math]\frac{a^n+b^n}{a+b}[/math], i.e. [math]W[/math] is NOT divided by [math]V[/math].

For proof lemma-theorem in genera case maybe one can use the fact that - in conditions of FLT – the numbers [math]a_2[/math] andи [math]b_2[/math] little differ one from other. There are many other considerations.

 

There are many other considerations.

 

Lemma-theorem (consequently and VTF) can be brought to the different simple theorems. Here one of them:

The numbers [math]\frac{a^n-b^n}{a-b}[/math] and [math]\frac{(a+1)^n-(b+1)^n}{a-b}[/math], where [math]a-b[/math] is not is divided by [math]n[/math] (and numbers [math]a[/math] and [math]b[/math] have no common factors), have no common factors.

What will say in regard to this the specialists in the theory of the numbers?

And... what will say schoolboys about actually five-line proof of FLT (or its reducing to lemma-theorem) in 8°-9°?

[uncool]

It's been a while, Victor, and I'm ready to try this again.

Complet text of the proof

 

 

All wholes numbers in the next are given in the prime base [math]n[/math].

 

Here are the known positions, repeatedly discussed on the mathematical forums. Thus, let

 

(1°) [math]A^n+B^n = C^n[/math], where prime [math]n > 2[/math] and [math]A, B, C[/math] have no common factors.

Let for the moment the number [math]ABC[/math] is not divided by [math]n[/math]. Then:

Looks alright to me so far.

 

(2°) [math]C^n-B^n=(C-B)P[/math], where [math]C-B=a'^n, P=a''^n[/math],

[math]C^n-A^n=(C-A)Q[/math], where [math]C-A=b'^n, Q=b''^n[/math],

[math]A^n+B^n=(A+B)R[/math], where [math]A+B=c'^n, R=c''^n[/math],

[this follows from the simple lemma: if the numbers [math]A[/math] and [math]B[/math] have no common factors and the number [math]A+B[/math] is not divided by [math]n[/math], then the numbers [math]A+B[/math] and [math]R = \frac{A^n+B^n}{A+B}[/math] have no common factors (since [math]R=(A+B)^2T+nAB)[/math]];

Ahem. A = 1, B = 2, n = 3. Thank you.

 

 

 

(3°) the numbers [math]a', a'', b', b'', c', c''[/math] have no common factors;

OK, let's get to work on this slowly. When you say no common factors, do you mean for all 6 numbers in total? or pairwise?

 

(4°) [math]A=a'a'', B=b'b'', C=c'c''[/math].

Sure.

 

(5°) Important equalities:

[math]c'^n - b'^n = 2a'a'' - (C-B)[/math],

[math]c'^n - a'^n = 2b'b'' - (C-A)[/math],

[math]a'^n + b'^n = 2c'c'' - (A+B)[/math], or

 

(5°a) [math](A+B) - b'^n = 2a'a'' - a'^n [/math],

[math](A+B)- a'^n = 2b'b'' - b'^n [/math],

[math]a'^n + b'^n = 2c'c'' - (A+B)[/math],

from here we find the numbers [math]a''[/math] and [math]b''[/math]:

Could you please show where you got this math? I have no idea where you got it from.

(6°) [math]a'' = \frac{(A+B) - b'^n + a'^n}{2a'}[/math],

[math]b'' = \frac{(A+B) - a'^n + b'^n}{2b'}[/math].

Assuming what you said in 5 is correct, then sure.

At last, we find two important formulas, лежащие in the base of the proof of the FLT:

(7a°) [math]a'' - b'' = \frac{(b'-a')(A+B) + (a'+b')(a'^n-b'^n)}{2a'b'}[/math],

(7b°) [math]a'' + b'' = \frac{(b'+a')(A+B) - (a'-b')(a'^n-b'^n)}{2a'b'}[/math].

Sure, works for me, at the moment.

That is my analysis so far. I've found an error and a few points I'd like you to expand upon. I will continue after you have done so.

=Uncool-

[Victor Sorok]

It's been a while, Victor, and I'm ready to try this again.=Uncool-

Note:

In new texts [math]A=a, B=b, C=c, a'=a_1, a''=a_2[/math], etc.

 

OK, let's get to work on this slowly. When you say no common factors, do you mean for all 6 numbers in total? or pairwise?

It is not true if, for example, c is divided by n:

Then the numbers [math]a+b[/math] and [math]R[/math] are divided by [math]n[/math] and the formulas 2° for [math]a+b[/math] and [math]R[/math] is not true.

 

Could you please show where you got this math? I have no idea where you got it from.

From 2°:

[math] a_1^n =c-b, P=a_2^n[/math],

[math] b_1^n =c-a, Q=b_2^n[/math],

[math] c_1^n =a+b, R=c_2^n[/math]:

[math] a_1^n +b_1^n=(c-b)+(c-a)=2c-(a+b)=2c_1c_2-c_1^n[/math], etc.

[uncool]

I'm still trying to find what you mean by no common factors.

Does that mean that every pair of numbers is relatively prime? or that no number besides 1 divides all 6?

 

Also, you haven't looked at the error I found.

=Uncool-

[Victor Sorok]

I'm still trying to find what you mean by no common factors.

1. Does that mean that every pair of numbers is relatively prime?

or that no number besides 1 divides all 6?

 

2. Also, you haven't looked at the error I found.

=Uncool-

1. Yes, every pair of numbers is relatively prime.

2. Please repeat the error which you found.

 

What will say in regard to this the specialists in the theory of the numbers?

In the march after the lemma!

 

I until did not find the simple proof of lemma. But here one of the ideas of its proof under the conditions VTF.

Is created impression, that remainder from the division of the number [math]a_2[/math] into [math]V[/math] less [math]a_1[/math], and remainder from the division of the number [math]b_2[/math] into [math]V[/math] less [math]b_1[/math]. IF THIS IS SO, then the truth of lemma is obvious.

[uncool]

You said you had a lemma where (A + B, (A^n + B^n)/(A + B)) = 1, where (a, b) means gcd(a, b). But if A = 1, B = 2, and n = 3, then you have a problem. I am sure I could find several more problems than that.

=Uncool-

[Victor Sorok]

You said you had a lemma where (A + B, (A^n + B^n)/(A + B)) = 1, where (a, b) means gcd(a, b). But if A = 1, B = 2, and n = 3, then you have a problem. I am sure I could find several more problems than that.

=Uncool-

In the Fermat’s equality all numbers a, b, c more then n^3.

[uncool]

Then a corresponding example would be:

A = 28, B = 29, n = 3.

A is equivalent to n mod 9, so A^3 will be the same mod 9, as will B^3. Then A+B will also be equivalent, and they will have a common factor of 3.

Also, with n = 5:

A = 127, B = 128. The common factor will be 3.

=Uncool-

[Victor Sorok]

Then a corresponding example would be:

A = 28, B = 29, n = 3.

A is equivalent to n mod 9, so A^3 will be the same mod 9, as will B^3. Then A+B will also be equivalent, and they will have a common factor of 3.

Also, with n = 5:

A = 127, B = 128. The common factor will be 3.

=Uncool-

 

You are right, but this does not refer to the proof:

A = 28 and B = 29 have no common factor.

++++++++

Unfortunately, it was possible to prove the lemma only on the condition that the numbers [math]a_2[/math] and [math]b_2[/math] have one (to Maxim two) prime base, what it is clearly insufficient for the proof FLT. Possibly, something can give the use of the fact that each prime cofactor of the numbers [math]V, a_2[/math] and [math]b_2[/math] take the form [math]en+1[/math]…

[Victor Sorok]

THE HYPOTHETICAL PROOF OF FERMAT'S LAST THEOREM,

based on the following two lemma-theorems

 

Lemma 1:

For any odd number [math]a[/math] there is an infinite set of such prime numbers [math]q>a[/math], that the numbers [math]a[/math] and [math]q-1 [=m][/math] are mutually prime.

Lemma 2:

All last digits in the numbers [math]a^t[/math] in the base [math]q[/math], where [math]t=1, 2, … q-1[/math], are different, i.e., compose the complete set of positive digits in the base [math]q[/math].

[Now there is no time to recall the proofs of these lemmas - this can be put off to the future. But it is worthwhile to look comments of “shwedka” and “tolstopuz” on the page http://lib.mexmat.ru/forum/viewtopic.php?t=8525&start=150 (in Russian).]

 

Essence of the obtained contradiction of the FLT: in the base [math]q[/math] the number [math]a^n+b^n -c^n=1[/math], but not [math]0[/math].

 

Thus, let us assume that

(1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and [math]a[/math] or [math]c[/math] also odd.

 

Proof VTF

(2°) for the odd number [math]3na^n[/math] (or for [math]nc^n[/math]) let us take the prime number [math]q>3na^n>c^n[/math], which satisfies the condition of lemma 1.

Let us compile linear diophantus equation with the mutually prime parameters [math]n[/math] and [math]m=q-1[/math]:

(3°) [math]n(x+tm)-m(y+tn)=1[/math] with the general solution [math](x+tm; y+tn)[/math].

(4°) Since according to lemma 2 (since one-digit number [math]a[/math] and number [math]m[/math] mutually prime), all one-digit ends in [math]t-1[/math] numbers [math]a^t[/math] - consequently, and the numbers [math]a^x+tm[/math] - are different, there are such values [math]x+tm=r[/math] and [math]x+tm=s[/math] that the one-digit ends in the numbers [math]a^r[/math] and [math]a^s[/math] are equal to [math]b[/math] and [math]c[/math].

But according to Little Fermat's theorem, one-digit ends in the numbers [math]a^n[/math], [math]a^{rn}[/math] and [math]a^{sn}[/math] in the base [math]q[/math] are equal to [math]1[/math], and then the one-digit end of the number

(5°) [math]a^n+a^{rn}-a^{sn}[/math], i.e., [math]a^n+b^n-c^n=1[/math], but not [math]a^n+b^n-c^n=0[/math].

 

FLT is proven.

[Victor Sorok]

THE HYPOTHETICAL PROOF OF FERMAT'S LAST THEOREM...

 

News

 

Well and now - the most essential COMPUTER fact in a matter of elementary proof FLT:

On the set [math]N[/math] of last digits in the numbers [math]q^n[/math], where [math]q[/math] is prime and [math]n [>2][/mat] is a divisor of the number [math]q-1[/math], the equality [math]A+B=C[/math] only if [math]A=1[/math] (or [math]B=1[/math]) or [math]A=B[/math] .

(Obviously, if prime [math]q>c^n[/math], then Fermat’e equality is impossible on the set [math]N[/math].)

There remains only to explain to us computer fact.

[uncool]

THE HYPOTHETICAL PROOF OF FERMAT'S LAST THEOREM,

based on the following two lemma-theorems

 

Lemma 1:

For any odd number [math]a[/math] there is an infinite set of such prime numbers [math]q>a[/math], that the numbers [math]a[/math] and [math]q-1 [=m][/math] are mutually prime.

This is easy to prove using Dirischlet's theorem.

Lemma 2:

All last digits in the numbers [math]a^t[/math] in the base [math]q[/math], where [math]t=1, 2, … q-1[/math], are different, i.e., compose the complete set of positive digits in the base [math]q[/math].

Are we assuming that a is a generator of q? Because in general this is not true.

[Now there is no time to recall the proofs of these lemmas - this can be put off to the future. But it is worthwhile to look comments of “shwedka” and “tolstopuz” on the page http://lib.mexmat.ru/forum/viewtopic.php?t=8525&start=150 (in Russian).]

 

Essence of the obtained contradiction of the FLT: in the base [math]q[/math] the number [math]a^n+b^n -c^n=1[/math], but not [math]0[/math].

 

Thus, let us assume that

(1°) [math]a^n+b^n = c^n[/math], where odd [math]n >2[/math] and [math]a[/math] or [math]c[/math] also odd.

 

Proof VTF

(2°) for the odd number [math]3na^n[/math] (or for [math]nc^n[/math]) let us take the prime number [math]q>3na^n>c^n[/math], which satisfies the condition of lemma 1.

Which is it? do you want q greater than both?

Let us compile linear diophantus equation with the mutually prime parameters [math]n[/math] and [math]m=q-1[/math]:

(3°) [math]n(x+tm)-m(y+tn)=1[/math] with the general solution [math](x+tm; y+tn)[/math].

Since m and n are relatively prime, sure.

(4°) Since according to lemma 2 (since one-digit number [math]a[/math] and number [math]m[/math] mutually prime), all one-digit ends in [math]t-1[/math] numbers [math]a^t[/math] - consequently, and the numbers [math]a^x+tm[/math] - are different, there are such values [math]x+tm=r[/math] and [math]x+tm=s[/math] that the one-digit ends in the numbers [math]a^r[/math] and [math]a^s[/math] are equal to [math]b[/math] and [math]c[/math].

What? Could you please try to write this part again? I can't understand what you are saying at all.

But according to Little Fermat's theorem, one-digit ends in the numbers [math]a^n[/math], [math]a^{rn}[/math] and [math]a^{sn}[/math] in the base [math]q[/math] are equal to [math]1[/math], and then the one-digit end of the number

Are you saying that because you think a^n = 1 mod q?

(5°) [math]a^n+a^{rn}-a^{sn}[/math], i.e., [math]a^n+b^n-c^n=1[/math], but not [math]a^n+b^n-c^n=0[/math].

 

FLT is proven.

Why do a^{rn} and a^{sn} substitute for b^n and c^n?

=Uncool-

[Victor Sorok]

1. Are we assuming that a is a generator of q? Because in general this is not true.

2. Could you please try to write this part again? I can't understand what you are saying at all.

3. Are you saying that because you think a^n = 1 mod q?Why do a^{rn} and a^{sn} substitute for b^n and c^n?

=Uncool-

 

1. Yes.

2. Correction:

"(4°) Since according to lemma 2 (since one-digit number [math]a[/math] and number [math]m[/math] mutually prime), all one-digit ends in [math]t-1[/math] numbers [math]a^t[/math] - consequently, and the numbers [math]a^{x+tm}[/math] - are different, there are such values [math ]x+tm=r[/math] and [math ]x+tm=s[/math] that the one-digit ends in the numbers [math]a^r[/math] and [math]a^s[/math] are equal to [math]b[/math] and [math]c[/math]".

But "consequently, and the numbers [math]a^{x+tm}[/math]" is not true.

3. This is not true.

 

Thanks

[uncool]

Well, my first objection so far is that you are assuming that a is a generator for some large q. How do you know that there is such a q? This cannot be proven using Dirischlet's theorem.

 

And what do you mean by "this is not true" as a response to number 3? Are you saying that I read it wrong, or that you wrote it wrong?

=Uncool-

[Victor Sorok]

And what do you mean by "this is not true" as a response to number 3? Are you saying that I read it wrong, or that you wrote it wrong?

=Uncool-

 

I wrote incorrectly.

 

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

 

Position until today

 

Designation 1: [math]a_1[/math] - the last digit of any number [math]a[/math].

Here is the enumeration of the assertions, which are necessary for the proof FLT:

1) for the given ones of the prime number [math]n>2[/math] and the odd number [math]d[/math] there is an infinite set of prime numbers [math]q[/math] of the form [math]q=2pn+1[/math], where odd [math]p[/math] and [math]d[/math] are relatively prime.

For the proof FLT we will take

2) the number [math]d[/math], which is been the greatest odd divisor of the number [math](abc)^n[/math] from the Fermat’s equality, and

3) the number [math]q>c^n[/math] (i.e. the numbers [math]a^n, b^n, c^n[/math] are one-digit).

 

Designation 2: [math]N(q)[/math] – the set of different (without the alliterations) last digits in the numbers [math]t^n[/math] ([math]t=1, 2... q-1[/math]) in the prime base [math]q[/math]. Then

 

4)The number of different elements in the set [math]N(q)[/math] is equal [math](q-1)/n[/math].

5) Set [math]N(q)[/math] contains number [math]2[/math].

6) One half of set [math]N(q)[/math] is described by formula [math]2^t_1[/math] ([math]t=1, 2... (q -1)/(2n)[/math]), another half of set [math]N(q)[/math] is described by formula [math](q-2^t_1)_1[/math] ([math]t=1, 2... (q -1)/(2n)[/math]).

Two examples of [math]N(q)[/math] for [math]n=3[/math]:

[math]N(31)[/math]: 1, 2, 4, 8, 15, 16, 23 (=31-8), 27 (=31-4), 29 (=31-2), 30 (=31-1);

[math]N(43)[/math]: 1, 2, 4, 8, 11 (=43-32), 16, 21 (=64-43), 22 [=43-21=43-(64-43)], 27 (=43-16), 32, 35 (=43-8), 39 (=43-4), 41 (=43-2), 42 (=43-1).

 

Thus, in the base [math]q[/math] Fermat’s equality takes the interesting form:

7) [math](2^u_1 + 2^v_1-2^w_1)_1=0[/math] (or even [math](1 + 2^v_1-2^w_1)_1=0[/math]), where [math]2^u_1=a^n, 2^v_1=b^n, 2^w=c^n[/math].

However, it can prove to be insufficiently for the completion of the proof of my knowledge. I propose to you to complete proof FLT.

 

Time out

[uncool]

How do you know that 2 is an nth power? And how do you know that its powers generate half of N(q)?

And where do you use d?

And what is the problem with the final form that you have written?

=Uncool-