Proof of FLT in the binary system

[Victor Sorok]

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.

[insane_alien]

Detailed calculations will be published in proportion to their readiness.

 

For those that don't have to justify timescales to a boss/advisor/evil overlord this means they'll be posted after he's actually done the calculations.

[Victor Sorok]

For those that don't have to justify timescales to a boss/advisor/evil overlord this means they'll be posted after he's actually done the calculations.

 

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.

[Victor Sorok]

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.

 

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

[uncool]

How is D equal to A^n + B^n - C^n?

 

Why can't A^n + B^n - C^n have an extra 0 at the end?

 

A - a = dn^(k+1) - 2a, not dn^(k+1), and so on. Perhaps you meant to write D + E?

 

Can you try to write this all out formally, starting from the beginning? It's a bit hard to read it all bunched up as you have it.

=Uncool-

[Victor Sorok]

How is D equal to A^n + B^n - C^n?

 

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.

[EvoN1020v]

Victor Sorok, do you mind using Latex next time? No offence, but your posts are full of scribbles which means it's hard to read your assumed ideal proof. Thank you.

[uncool]

Can you prove the lemma? I'm not sure that it's true. In fact, if it were true, it would prove FLT all by itself, because:

 

If a^n + b^n - c^n = 0 = nabcd, then either n = 0 (not true), a, b, or c = 0 (assumed not true), so d = 0, but then n|d, which you don't allow.

 

Also, do you mean (c-a)^n + (c-b)^n - (a+b)^n?

 

How do you know that (c-a), (c-b), and (a+b) have no factors of n? If they do have enough factors of n, then they can be equal.

[Victor Sorok]

Victor Sorok, do you mind using Latex next time?

 

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°)!!!

[Victor Sorok]

Can you prove the lemma? I'm not sure that it's true.

 

I had lost.

 

But cf. Splendid joke.

 

Thanks

[Cap'n Refsmmat]

Excuse me last time, please!

 

Splendid joke.

He meant to use the [math] tags.

[TriggerGrinn]

What is it you are refering to with "FLT"

[Dave]

What is it you are refering to with "FLT"

 

Fermat's Last Theorem.

[Ragib]

O MY JESUS YOUR JOKING. I thought he meant something else, anything else! I didnt actually read the posts seeing as I didnt know what FLT was...but omg Fermats Last Theorem? Proved, so easily, When The Best mathematicians in the field of modular forms would have trouble UNDERSTANDING Andrew Walters 200 page proof. Get some Humility.

[uncool]

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°)!!!

 

I do not see 2 as being true. 3 either. Are you assuming n is prime?

=Uncool-

[uncool]

O MY JESUS YOUR JOKING. I thought he meant something else, anything else! I didnt actually read the posts seeing as I didnt know what FLT was...but omg Fermats Last Theorem? Proved, so easily, When The Best mathematicians in the field of modular forms would have trouble UNDERSTANDING Andrew Walters 200 page proof. Get some Humility.

 

This is why he is asking and correcting...it is possible, though unlikely, that he's found it.

Side note: Andrew Wiles.

There is nothing wrong in asking whether this will work...

=Uncool-

[Ragib]

Ok im sorry, just found it kinda shocking. And of course, what was I thinking..Andrew Walters...ffs ty

[Victor Sorok]

It is seemed that I found the simple proof of FLT in the binary system.

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.

[Victor Sorok]

Elementary proof of Fermat's last theorem in base [MATH]3[/MATH] for [MATH]n>3[/MATH] (numbers [MATH]a, b, c[/MATH] mutually-prime).

 

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

[Victor Sorok]

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).

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.

[Victor Sorok]

I return...

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.

[Victor Sorok]

Interesting lemma

 

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 ".

[Victor Sorok]

Interesting lemma

 

Here is the Lemma, which easily proves and which is sufficient for the brief proof of the FLT:

Lemma. If integers a, b, c have only one common divisor 1, a+b=c, number (c^n-a^n) and (c^n-b^n) have common prime divisor d>3c^2, in base d numbers a^n, b^n, c^n finish by digit 1, then with n>1 either number (a^n-b^n)/(a+b) is not divided by d, or number 3c^2 is divided into d ".

 

to be continued

[Victor Sorok]

to be continued

 

Lemma.

If integers [MATH]a, b, c[/MATH] have only one common divisor [MATH]1[/MATH], [MATH]a+b=c[/MATH], number [MATH](c^n-a^n)[/MATH] and [MATH](c^n-b^n) [/MATH] have common prime divisor [MATH]d>3c^2[/MATH], in base d numbers [MATH]a^n, b^n, c^n[/MATH] finish by digit [MATH]1[/MATH], then with [MATH]n>1[/MATH] either number [MATH](a^n-b^n)/(a+b) [/MATH] is not divided by [MATH]d[/MATH], or number [MATH]3c^2[/MATH] is divided by [MATH]d[/MATH]d.

 

Proof.

Let us assume that number [MATH](a^n-b^n)/(a+b)[/MATH] is divided by d.

Then number [MATH]S=(c^n-b^n)/(c-b)+(c^n-a^n)/(c-a)+(a^n-b^n)/(a+b)[/MATH], or

[MATH] (c^n-b^n)/a+(c^n-a^n)/b+(a^n-b^n)/c = [c^(n+2)+b^(n+2)+(a+b)c^(n+1]/abc=[/MATH]

[MATH]=[c^2+b^2+a(c+b)]/abc[/MATH], is divided by [MATH]d[/MATH].

But since number [MATH]a^n, b^n, c^n[/MATH] finish by number [MATH]1[/MATH], then the number [MATH]s = c^2+b^2+a(c+b)[/MATH], where [MATH]0<s<d[/MATH], is single-digit number and, therefore, number [MATH]S[/MATH] is NOT DIVIDED by [MATH]d[/MATH].

[Victor Sorok]

Lemma.

If integers [MATH]a, b, c[/MATH] have only one common divisor [MATH]1[/MATH], [MATH]a+b=c[/MATH], number [MATH](c^n-a^n)[/MATH] and [MATH](c^n-b^n) [/MATH] have common prime divisor [MATH]d>3c^2[/MATH], in base d numbers [MATH]a^n, b^n, c^n[/MATH] finish by digit [MATH]1[/MATH], then with [MATH]n>1[/MATH] either number [MATH](a^n-b^n)/(a+b) [/MATH] is not divided by [MATH]d[/MATH], or number [MATH]3c^2[/MATH] is divided by [MATH]d[/MATH]d.

+++

Elementary proof of Fermat's last theorem

 

According to known theorem from the theory of integers, for any prime [MATH]n[/MATH] there exist infinite great number of the prime numbers [MATH]d[/MATH] of form [MATH]d=pn+1[/MATH], where [MATH]p[/MATH] is integer. (This theorem is the simple consequence from the fact that, if numbers [MATH]a[/MATH] and [MATH]b[/MATH] are mutually-prime and [MATH]n[/MATH] prime, then each divisor of the number [MATH] (a^n+b^n)/(a+b)[/MATH] takes form [MATH]pn+1[/MATH]).

 

Proof FLT:

Let us take simple [MATH]d>c^{4n}[/MATH]. Then, according to Fermat's little theorem, the numbers [MATH]C^p-B^p, C^p-A^p, A^p-B^p[/MATH], where [MATH]C=c^n, B=b^n, A=a^n[/MATH], are divided by [MATH]d[/MATH]. And since [MATH]A+B=C[/MATH], then, according to Lemma, number [MATH]3C^2[/MATH] is divided by [MATH]d>3C^2[/MATH]. Consequently, the solution of Fermat’s equation is not integer.