My Math Forum FLT n=3 means A divisible by (C-B) ?

 Number Theory Number Theory Math Forum

 April 23rd, 2016, 05:13 AM #11 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Luckily my last post was blown away for an unknown reason.... The point is show that a proportional condition on $Y$ (here Delta/Delta=1) on a curve $Y=X^2$, that has a first linear derivate can again "generate" a proportional condition on $X$, here $(C-A)/(C-B)= p/q$ with $(p,q)\in N$. While this cannot happen on $Y=X^n$ wiht $n>2$ because the non linear first derivates define an irrational ratio between the "middle" integer height ($C^n\in N$) and the correspondent irrational X ($C$), that is his n-th root. Math asap.... but picture already tell the story... Happy to hear that Mr. Whiles becomes Sir. and 750KUsd reach, while we, humans, are already searching for a more usefull solution to FLT problem ;-P Last edited by complicatemodulus; April 23rd, 2016 at 05:18 AM.
 May 14th, 2016, 11:39 PM #12 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I return in math remembering my first post in this topic: Part 1) FLT rewritten with Sum is: $\displaystyle A^3 = \sum_{X=B+1}^{C} (3X^2-3X+1)$ or $\displaystyle A^3 = \sum_{X=1}^{C-B} (3(X+B)^2-3(X+B)+1)$ or $\displaystyle A^3 = \sum_{X=1}^{C-B} (3X^2+6BX+3B^2-3X-3B +1)$ or $\displaystyle A^3 = \sum_{X=1}^{C-B} ((3X^2-3X +1) +6BX-3B +3B^2)$ or $\displaystyle A^3 = (C-B)^3 + 3B \sum_{X=1}^{C-B} (2X-1) +3B^2(C-B)$ or $\displaystyle A^3 = (C-B)^3 + 3B (C-B)^2 +3B^2(C-B)$ so $(C-B)>=1$ is a factor of A Part 2) for the same reason/process will be: $\displaystyle B^3 = (C-A)^3 + 3A (C-A)^2 +3A^2(C-A)$ so $(C-A)>=1$ is a factor of B This is true for any $n>=2$ since rising n the limits are the same and change just the term $Mn= (X^n-(X-1)^n)$ Part 3) Why for n=2 there is a solution: Remembering via Step Sum we prove we are able to represent also Rational value/powers, for n=2 we can write: (1) $\displaystyle \sum_{X=1}^{A} (2X-1) = \sum_{X=B+1}^{C} (2X-1)= \sum_{X=1}^{C-B} (2(X+B)-1)$ Introducing the Step Sum, keeping: $K=A$ and calling $x=X/A$ I already show nothing change if we will rewrite the (1) as the Step Sum, Step $x_1=1/A$, $x_2= 2/A$; $x_3= 3/A$ etc...: $\displaystyle \sum_{x=1/A}^{A} (2x/A-1/A^2) = \sum_{x=1/A}^{C-B} (2(x+B)/A-1/A^2))$ And now we can divide both terms by $A^2$, dividing the upper limit of the Step Sum by $A$: $\displaystyle \sum_{x=1/A}^{1} (2x/A-1/A^2) = \sum_{x=1/A}^{(C-B)/A)} (2(x+B)/A-1/A^2))$ But remembering the is: $A= \pi_1 *(C-B)$ And solving and grouping the new known square: $\displaystyle 1= \sum_{x=1/A}^{(C-B)/A)} (2x/A-1/A^2)) + 2B/A *((C-B)/A)$ or: $\displaystyle 1 = ((C-B)/A)^2 + 2B(C-B)/(A^2)$ or: $\displaystyle 1 = 1/{\pi_1}^2 + 2B/((C-B)* \pi_1)$ or: $\displaystyle {{\pi_1}^2} * (C-B) = (C-B)+2B$ or: $\displaystyle {\pi_1}^2 = (C+B)/(C-B)$ that fit for many values: f.ex $\displaystyle {\pi_1}^2 = (5+4)/(5-4) = 9$ $\displaystyle {\pi_1}=A = 3$ As already written it works for any $A>=3$ infact in case "A" is ODD: $A$; $B= (A^2-1)/2$ ; $C=(A^2+1)/2$ in case "A" is EVEN: $A$; $B= (A/2)^2 -1)/2$ ; $C=(A/2)^2+1$ Part 4) Why Fermat is right for $n=3$ Because doing the same tedious "process" starting from: $\displaystyle A^3 = \sum_{X=B+1}^{C} (3X^2-3X+1)$ Sorry for this mental torture... I'll jump directly to the reduced Step Sum already divided by $A^3$: $\displaystyle \sum_{x=1/A}^{1} (3x^2/A-3x/A^2+ 1/A^3) = \sum_{x=1/A}^{(C-B)/A} (3(x+B)^2/A -3(x+B)/A^2+1/A^3))$ so.... remembering the is: $A= \pi_1 *(C-B)$ or $B= \pi_2 *(C-A)$ we arrive to the non elegant solution: $\displaystyle {\pi_1}^3- 3B^2{\pi_1}/(C-B) - 3B-1 =0$ or (if we start from $B^3=C^3- A^3$) to: $\displaystyle {\pi_2}^3- 3A^2{\pi_1}/(C-A) - 3A-1 =0$ that is not elegant AT ALL since rising n we have again to repeat all this torture... ...but I hope here is much easy to "see" what is the dust (what I call the mixed product) that lock the gears rising n... Waiting for your kindly check... and probably a way to add some shortcut using your more powerfull math... Thanks Ciao Stefano Last edited by complicatemodulus; May 14th, 2016 at 11:44 PM.
 May 15th, 2016, 02:01 AM #13 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 One minute more on, let me have an hope for the final elegant proof just remembering the binomial develope rule... Thanks Ciao Stefano
May 15th, 2016, 09:21 AM   #14
Senior Member

Joined: Sep 2010

Posts: 221
Thanks: 20

Quote:
 Originally Posted by complicatemodulus FLT n=3 means A divisible by C-B ?

A^n=(C-B)[C^(n-1)+....+B^(n-1)]=(C-B){(C-B)^(n-1)+nCB[C^n-3)-...+B^(n-3)]}

C-B and polynomial in brackets are coprime unless C-B is divided by n. The rest of its factors must be to the power n (including n=3) and divide A.

May 15th, 2016, 08:51 PM   #15
Banned Camp

Joined: Dec 2012

Posts: 1,028
Thanks: 24

Quote:
 Originally Posted by McPogor Without touching upon your considerations A^n=(C-B)[C^(n-1)+....+B^(n-1)]=(C-B){(C-B)^(n-1)+nCB[C^n-3)-...+B^(n-3)]} C-B and polynomial in brackets are coprime unless C-B is divided by n. The rest of its factors must be to the power n (including n=3) and divide A.
Thanks, that was very clear... but the point is very far from the the topic's title...

 Tags divisible, flt, means

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post JohnnyGs Linear Algebra 5 October 18th, 2015 04:46 AM Rama Number Theory 2 February 2nd, 2014 01:30 PM jasonjason Computer Science 0 July 26th, 2013 01:47 AM swat532 Real Analysis 3 September 15th, 2007 05:18 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top