My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree1Thanks
Reply
 
LinkBack Thread Tools Display Modes
December 19th, 2014, 02:21 AM   #1
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

Fermat the last...

Knowing FLT conditions , so A<B<C integers

Asking if C^-B^n = A^n

From the identity in R:

(1) $\displaystyle \int_{B}^{C}{ [n*x^{(n-1)}]} dx = \int_{0}^{C-B}{ [ n(x+B)^{(n-1)}]} dx = C^n-B^n $

I've shown how to re-write in N (see my previous posts)

So we check if there is a solution to the (1) in N (in the example n=3):

We know we can make and Exchange of variable calling

$\displaystyle X=(x+B) $


$\displaystyle \sum_{m =B+1}^{C} {(3m^2 -3m +1) } =? \sum_{X =1}^{C-B} {(3X^2 -3X +1)} $

or:

$\displaystyle \sum_{m =B+1}^{C} {(3m^2 -3m +1)} =? \sum_{x =1}^{C-B} {(3(x+B)^2 -3(x+B) +1)} $

or:

$\displaystyle C^3-B^3 =? \sum_{x =1}^{C-B} {(3x^2 -3x +1)} + 3B \sum_{x =1}^{C-B} (x) + (C-B)*(B^2-3B+1) $

or:

$\displaystyle C^3-B^3 =? (C-B)^3 + 3B (C*(C+1)/2) + (CB^2-3BC+C-B^3-3B^2+B) $


or:

$\displaystyle A^3 =? C^3-B^3 =? C^3-B^3-3{C^2} B+3CB^2 + {(3/2)}BC^2 +{(3/2)}B +(CB^2 - 3BC + C - B^3 - 3B^2+B) $

etc.....


The same can be done in Q using my step sum:

$\displaystyle \sum_{x =B+1/k}^{C}(s) {[3x^2/k-3x/k^2+1/k^3]} =? \sum_{x =1/k}^{C-B} {(3(x+B)^2 /k -3(x+B)/k^2 +1/k^3)} $


...No way to find a triplet A, B,C under Fermat conditions...

Thanks
Ciao
Stefano

Last edited by complicatemodulus; December 19th, 2014 at 02:59 AM.
complicatemodulus is offline  
 
December 19th, 2014, 07:34 AM   #2
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 932

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
Originally Posted by complicatemodulus View Post
...No way to find a triplet A, B,C under Fermat conditions...
Everything else in your post aside, you don't demonstrate this.
CRGreathouse is offline  
December 19th, 2014, 07:46 AM   #3
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

Sorry I'm not Einstein...

It's just on the way for... you probably can check, correct, and perhaps conclude.


Thanks
Ciao
Stefano
complicatemodulus is offline  
December 19th, 2014, 08:06 AM   #4
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

With n=3, C=B+1 seems doesen't works, so no integer B satisfy the identity.

I hope the rest will came...

Thansk
Ciao
Stefano
complicatemodulus is offline  
December 22nd, 2014, 07:44 AM   #5
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

Here the actual revision on the Sum / Step sum / n-th problems:

http://www.maruelli.com/COMPLICATE%2...tion%20def.pdf

There are still many errors but I hope sum interesting point on.

Thanks
Ciao
Stefano
complicatemodulus is offline  
December 24th, 2014, 09:58 AM   #6
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

Same link, I correct this wrong graph:



Thanks & Merry Christmass !
Ciao
Stefano
complicatemodulus is offline  
December 27th, 2014, 09:10 AM   #7
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

If I've not made some mistake in the develope:

$\displaystyle -3{C^2} B+3CB^2 + {(3/2)}BC^2 +{(3/2)}B +(CB^2 - 3BC + C - B^3 - 3B^2+B) =0 $

Has No integers solutions.


-3*C^2*B+3*C*B^2+(3/2)*B*C^2+(3/2)*B+C*B^2-3*C*B+C-B^3-3*B^2+B=0 integer solutions - Wolfram|Alpha

And the same into the rationals:

-3*C^2*B+3*C*B^2+(3/2)*B*C^2+(3/2)*B+C*B^2-3*C*B+C-B^3-3*B^2+B=0 rational solutions - Wolfram|Alpha


Thanks
Ciao
Stefano
complicatemodulus is offline  
December 27th, 2014, 11:33 AM   #8
Global Moderator
 
CRGreathouse's Avatar
 
Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 932

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Of course W|A knows that FLT has already been solved by Wiles, and uses this to answer Diophantine queries.
CRGreathouse is offline  
December 27th, 2014, 11:25 PM   #9
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

Yes, always a round circle...

I'm working on the differential equation, but is hard for me...

Thanks
Ciao
Stefano
complicatemodulus is offline  
March 4th, 2015, 06:14 AM   #10
Senior Member
 
Joined: Dec 2012

Posts: 930
Thanks: 23

I forgot to write this to clarify my complicate-modulus conclusion on the FLT:

Knowing FLT conditions n>=3 odd, so A<B<C integers

Asking if

Is equal to ask if (here the example n=3, but is the same if n>3):

(1)

But I know, also, that:



And this is equal to:



So once you shift the right hand sum (1) to lower limit =1 (as already posted here)

So you can apply the same division on the last term. The result must return an integer, if the sum is equal to the cube A^3.

Than FLT follows for parity check on A,B,C on the divisibility of the new upper limit by A, in both case A,C ODDs, or A,C Evens.


Thanks
Ciao
Stefano

Last edited by skipjack; March 5th, 2015 at 11:52 AM.
complicatemodulus is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
fermat



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Fermat's Last again McPogor Number Theory 0 July 28th, 2014 10:58 AM
Fermat primes fafa Number Theory 4 July 10th, 2013 11:25 AM
Fermat's Curve mathbalarka Algebra 2 August 5th, 2012 07:22 PM
Fermat fakesmile Calculus 0 June 2nd, 2009 05:36 AM
Fermat circum Number Theory 1 June 15th, 2008 05:20 PM





Copyright © 2017 My Math Forum. All rights reserved.