## Fermat's Last Theorem

Math help on Congruences, Modulo, Fermat's Theorem, Fermat's Last Theorem (FLT), Euler's Theorem, Euler Totient Function, Divisors, Multiples, Prime Numbers, Prime Number Theorem, Distribution of Prime Numbers, Dzeta Function, Riemann Hypothesis, Algebraic Number Theory, Analytic Number Theory on My Math Forum.
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### Re: Fermat's Last Theorem

CRGreathouse wrote:http://xkcd.com/955/

Thanks!

+++

A simple proof of Fermat's Last Theorem

In Memory of mom

The idea of a complete proof of the WTF

The number T in equation $(C-B)^n+(C-A)^n=[(C-B)+(C-A)]T$ is divisible and is not divisible by the divisor r of C [=cr], which is not a divisor of A+B [=c^n].

The simplest (and not at all interesting) school justification of this fact (without actually computing!) will be presented on mathforums after the publication. (Anyone who knows the properties of the Fermat’s equation, can easily find justification for their own.)

So, before the last meeting.

January 13, 2012. Mezos
victorsorokin
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### Re: Fermat's Last Theorem

2-nd theorem on divisors of binomial power.

1-st (proved) theorem states:
In the equation $A^n+B^n=(A+B)R$, where the prime n>2 and integers A and B are relatively prime, every prime factor m (except n) of R has the form $m=dn+1$.

Also proved (Theorem 1a) that if the base A and B are n-th power, then
every prime factor m (except n) of R has the form $m=dn^2+1$.

2-nd (may not be proven) the theorem states:
If the bases A and B are not n-th power, then
number R contains a prime divisor m of the form $m=dn+1$, where d is not divisible by n.

For the existence of an elementary half page proof of the FLT missing only the 2nd theorem (perhaps proven).

The problem posed. Who decides it?
Last edited by The Chaz on Sun Jan 22, 2012 7:41 pm, edited 1 time in total.
Reason: Seriously, with the super-sized text?
victorsorokin
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### Re: Fermat's Last Theorem

2^3 + 5^3 = 7(3*6+1), but 6 is divisible by 3 and 3*6+1 is prime.
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/~bill/mingw/PARI-2-6.exe

CRGreathouse
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### Re: Fermat's Last Theorem

CRGreathouse wrote:2^3 + 5^3 = 7(3*6+1), but 6 is divisible by 3 and 3*6+1 is prime.

1) Thank!

2) You have: n=3 and 3*6+1=nd+1=m.

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

Victor Sorokin

A simple proof of Fermat's Last Theorem

In Memory of mom

By the end of the study

Today, January 24, 2012 two elementary proofs of the FLT were docked. (All the elements of proofs have been repeatedly published in mathforums.)

The proof consists of two parts:

1. Lemma: In the base case in the Fermat’s equality $C^n=A^n+B^n$ [$=(A+B)R=C'^n*R'^n=(C'R')^n$] every prime factor m (except n) of R' has the form $m=dn^2+1$. Consequently, two-digit ending of the number R' is equal to 01 and c^n≡c (mod n^2), where c is the last digit of the significand of C. From this it follows that

2. Penultimate digits in numbers 1^n, 2^n,... (n-1)^n is 0, and hence a two-digit ending number $S=1^n+2^n +...+(n-1)^n$ is the sum of an arithmetic progression $S'=1+2+...+(n-1)$, ie, the number d0, where d is not equal to zero, which contradicts the direct calculation of the end of S (it is equal to 00, which is evident from the grouping of terms in the sum of S into pairs: $S=[1^n+(n-1)^n]+[2^n+(n-2)^n]+$...).
victorsorokin
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### Re: Fermat's Last Theorem

A simple proof of Fermat's Last Theorem

In Memory of Mom

An elementary proof is based on a simple Lemma:
In a number system with base $Q=q^n$, where prime q>2 and q≠n, the last digits in all the degrees $1^n$, $2^n$,… $(Q-1)^n$ are different, ie constitute a complete set of digits.

Proof
Assume the contrary: the degrees $D^n$ and $E^n$, or, after dropping in them of the greatest common divisor F of the bases D and E, the numbers d^n and e^n (where 0<e<d<Q) end by equal digits. Then the number $d^n-e^n=(d-e)R$ is divisible by Q.
However, the number d-e is not divisible by Q, because d-e<Q, and the number d-e can not be divided even by q, since in this case the number R (which is known is relatively prime with the number d-e) is also divisible by q.
From this it follows that all the divisors q belong in the number R. But in this case the integral equation d^n-e^n=(d-e)R is impossible, since $d^n-e^n.

The proof of FLT (by contradiction).

Suppose that
1°) $A^n=C^n-B^n$ [=(C-B)P], where the number A (apparently, composite) is not multiple by n [otherwise, instead of A we take the number B] and it has a prime divisor q>2 (q≠n).

We write down the number C and B in the database Q:
$C=tQ+c$and $B=sQ+b$, where $Q=q^n$, 0<c<Q, 0<b<Q.

Then we rewrite 1° in the form:
2°) $UQ=(TQ+c^n)-(SQ+b^n)$, where, according to Lemma, in the base Q the numbers $c^n$ and $b^n$ end by different digits, and therefore, equation 2°, which is identically 1°, are contradictory.

Thus, the FLT is proved.

(Mezos, March 1, 2012)
victorsorokin
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### Re: Fermat's Last Theorem

victorsorokin wrote:Suppose that
1°) $A^n=C^n-B^n$ [=(C-B)P], where the number A (apparently, composite) is not multiple by n [otherwise, instead of A we take the number B] and it has a prime divisor q>2 (q≠n).

Technically this is inadmissible, as it does not handle the case
$(2^k)^n+(n\beta)^n=c^n.$

victorsorokin wrote:Lemma:
In a number system with base $Q=q^n$, where prime q>2 and q≠n, the last digits in all the degrees $1^n$, $2^n$,… $(Q-1)^n$ are different, ie constitute a complete set of digits.

False. Look at Q = 27, for example: 3^3 and 6^3 both end with the digit 0. 8^3 and 17^3 both end with the digit 26. Etc.
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/~bill/mingw/PARI-2-6.exe

CRGreathouse
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### Re: Fermat's Last Theorem

CRGreathouse wrote:False. Look at Q = 27, for example: 3^3 and 6^3 both end with the digit 0. 8^3 and 17^3 both end with the digit 26. Etc.

Actually this is not a counterexample to the lemma, because in the case $Q = 27$, $Q = 3^3$ and the lemma only handles the cases in which $q \neq n$.
icemanfan

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### Re: Fermat's Last Theorem

icemanfan wrote:
CRGreathouse wrote:False. Look at Q = 27, for example: 3^3 and 6^3 both end with the digit 0. 8^3 and 17^3 both end with the digit 26. Etc.

Actually this is not a counterexample to the lemma, because in the case $Q = 27$, $Q = 3^3$ and the lemma only handles the cases in which $q \neq n$.

Oops. Take Q = 81, then, and notice that 80^4 and 1^4 both end in 1 mod 81, or that 4^4 and 77^4 end in 13.
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/~bill/mingw/PARI-2-6.exe

CRGreathouse
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### Re: Fermat's Last Theorem

CRGreathouse wrote:
victorsorokin wrote:Lemma:
In a number system with base $Q=q^n$, where prime q>2 and q≠n, the last digits in all the degrees $1^n$, $2^n$,… $(Q-1)^n$ are different, ie constitute a complete set of digits.

False. Look at Q = 27, for example: 3^3 and 6^3 both end with the digit 0. 8^3 and 17^3 both end with the digit 26. Etc.

If Q=27, then q=3. But q≠n.
victorsorokin
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### Re: Fermat's Last Theorem

If Q=27, then q=3. But q≠n.
victorsorokin
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### Re: Fermat's Last Theorem

CRGreathouse wrote:Oops. Take Q = 81, then, and notice that 80^4 and 1^4 both end in 1 mod 81, or that 4^4 and 77^4 end in 13.

q=?
n=?
Is n prime?
Is q prime?
victorsorokin
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### Re: Fermat's Last Theorem

Clarification

In the proof of FLT (with a minimum value of C) as a working number we take the number of A or B, in which the factor C-B or C-A contains a divisor q, different from the numbers 2, n, dn+1. (Let me remind all the prime factors of large factors of X, Y, Z in the numbers of $A^n+B^n=(A+B)X, C^n-B^n=(C-B)Y, C^n-A^n=(C-A)Z$ have the form dn+1.)
Lemma is proved with this value of q.
victorsorokin
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### Re: Fermat's Last Theorem

victorsorokin wrote:
CRGreathouse wrote:Oops. Take Q = 81, then, and notice that 80^4 and 1^4 both end in 1 mod 81, or that 4^4 and 77^4 end in 13.

q=?
n=?
Is n prime?
Is q prime?

Let Q = 9. Then $1^2, 8^2$ both end in 1.
icemanfan

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### Re: Fermat's Last Theorem

icemanfan wrote:Let Q = 9. Then $1^2, 8^2$ both end in 1.

Right. There are lots of examples; victorsorokin's lemma fails about as hard as it could. In fact there aren't any examples with exponent > 1 where it works!
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/~bill/mingw/PARI-2-6.exe

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### Re: Fermat's Last Theorem

information

1. All previous proofs of the FLT, seems to be wrong.
2. Mistakes are useful because they show the right path to a solution.

3. Several years ago I proved (and published on the mathforums) two theorems:

A theorem on prime divisors of binomial power:
If the numbers A and B are relatively prime, and AB≠0 mod n, then every prime divisor m≠n of the number R in the equation $A^n+B^n=(A+B )R$, where a prime n>2 has a form m=dn+1.

If AB≠0 mod n then the number R in the equality $A^n+B^n=(A+B )R$ does not contain of prime divisors $m=dn^t +1$, where d≠0 mod n and t=0 or 1.

Today I found proof of this lemma for t>1, based on a theorem:

In a number system with base $p=2^{n^k}$ (n is prime) all the latest digits in the numbers
$1^n, 2^n,... (p-1)^n$ are different.

Thus, the number R contains no prime divisors of $m=dn^t+1$.

To be continued.
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