August 6th, 2017, 10:49 PM  #1 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  FLT from a system of infinite linear equations
FLT from a system of infinite linear equations As told from the Power properties they can be written as Sum: $$A^n= \sum_{x=1}^{A} M_n$$ $$B^n= \sum_{x=1}^{B} M_n$$ $$C^n= \sum_{x=1}^{C} M_n$$ Where: $$M_n=[x^n(x1)^n]$$ Than FLT state: $$C^n=A^n+B^n$$ Can be rewritten also as: $C^n=2A^n+ \Delta$ or: $C^n=2B^n \Delta$ Where: $$\Delta = \sum_{x=A+1}^{B} M_n$$ Than one can start (infinitely) to produce new equations summing $C^n$ each time: $2C^n = 3A^n+B^n + \Delta$ $2C^n = 3B^n+A^n  \Delta$ and from there go ahead in summations: $3C^n = 5A^n+B^n+ 2\Delta$ $3C^n = 5B^n+A^n 2\Delta$ but also: $4C^n = 7A^n+B^n+ 3 \Delta$ $4C^n = 7B^n+A^n 3 \Delta$ ...... so: $p * C^n = (2p1) A_p^n+B^n+ (p1) \Delta$ $p * C^n = (2p1) B^n+A^n (p1) \Delta$ To help: $C,B$ are known fixed Integers. $p$ we can (each time) decide who is, than we have to discuss about $A_p$, $n$ and $\Delta$... 
August 9th, 2017, 01:01 AM  #2 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
No interest on this ? But 20 pages on the wrong one...

August 9th, 2017, 04:05 AM  #3 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 305 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry 
I can't see anything wrong with what you have so far, but I can't see where you're trying to take this argument or how it could possibly help prove FLT...

August 9th, 2017, 04:48 AM  #4 
Senior Member Joined: Oct 2009 Posts: 733 Thanks: 246 
Is this supposed to prove FLT? What is this?

August 9th, 2017, 05:37 AM  #5 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
Thanks is 100% right, and from there you can try to argue somethink about the condition for $A$ or $B$ let this be possible. We can see that the left hand lineary rise, so also the right hand has to do the same. Supposing $C^n$ and $B^n$ be known constant than The first term in the right hand lineary rise (2 times faster), while the coefficient of the difference $\Delta$ that is $B^nA^n$ linearly rise (of the same ratio of the left hand 1) Since there must be a solution for each equation, seems to me that the only way is that the first derivate defining the area that rise from $A^n$ to $B^n$ (so $\Delta$) has to be linear too and with the known ratio $y'=2x$ or this term will rise faster than the others and there is no way to rearrange all the next equations... except going infimus. You agree on this ? 
August 9th, 2017, 07:42 AM  #6  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 305 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
 
August 9th, 2017, 07:57 AM  #7 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
I'm working on for having the support of a collection of numerical examples. Thanks Ciao Stefano Last edited by skipjack; August 9th, 2017 at 05:08 PM. 
August 9th, 2017, 04:32 PM  #8 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 305 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry 
If you express your argument clearly, people will be able to respond to it and give useful feedback. Until then, there's nothing valuable we can say, no matter how many examples you give.

August 9th, 2017, 07:26 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond 
This has been going on for years. He (complicatemodulus) talks about having solved FLT in an elementary sense but never explains his work sufficiently enough to lend any understanding to it. I've long ago given up involving myself in any of his threads as a result. 
August 9th, 2017, 09:57 PM  #10 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
Harsh and direct: Can you solve this linear system ?  Yes ! Pls let us know how (there are infinite more equations than variables...)  NO ? Pls avoid loose our time and talk about of "style"... or insult... Thanks... I never said in this topic this is a proof for FLT, I just said this is another trick FLT must satisfy; and if it works as it seems to me, probably, we have another "simple" proof. This is a forum where talk of math, so I post my conclusion to be checked AND my doubts... Thanks for thoose have the patience to read me, and try to correct my errors, or simply support with a "like"... The math comunity is fully ignoring my Step Sum from the beginning and this is quite old style stupid, but we know history never teach to be more carefull to new different thinking.... Ciao Stefano 

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