My Math Forum FLT proof

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 May 18th, 2015, 09:15 AM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 FLT proof Sorry I re-post cleaning as I can, hopeing someone agree with me that this is the FLT proof or what he, probably, call infinite descent (or he is looking for): Example for n=3 but nothing change for n>3 $A^3= C^3-B^3$ From FLT reduced to a sum: $\sum_{x=1}^{A} {(3x^2 -3x +1)} =? \sum_{ x=B+1}^{C} {(3x^2 -3x +1)}$ Now is possible to lower the right hand limits (without changing the result of it) as: $\sum_{x=1}^{A} {(3x^2 -3x +1)} =? \sum_{ x=1}^{C-B} {(3(x+B)^2 -3(x+B) +1)}$ 2) I transform both left and right hand terms in a Step Sum, step 1/K (as I've alredy shown) where is clear left hand rest A^n: (to do that I've made a change of variable from x to x= x*K living the same symbol to avoid loss of time in writing, but I've already shown that): $\sum_{x=1}^{A} {(3(x^2)/K -(3x)/K^2 +1/K^3)} =? \sum_{ x=1}^{C-B} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}$ --> AT THIS POINT WE CAN SHOW BY COMPUTATION THAT RISING "K" WE NEVER FIND A RESULT OR WE ALWAYS HAVE BACK : $\sum_{x=1}^{A} {(3(x^2)/K -(3x)/K^2 +1/K^3)} \neq \sum_{ x=1}^{C-B} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}$ AND WE ARE SURE FERMAT IS RIGHT SINCE IF, AND ONLY IF WE PASS TO THE LIMIT FOR K->INFINITY (SO VARIABLE IN R), WE HAVE AGAIN THE EQUALTY IS TRUE: 3) Make the limit with K-> infinity then pass to the definite integral. $\lim_{k\to\infty} { \sum_{x=1}^{A} {(3(x^2)/K -(3x)/K^2 +1/K^3)} = ? \lim_{k\to\infty}{ \sum_{ x=1}^{C-B} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}$ 4) Check if the definite integral (from 0 to ... in both side of two different terms) resulting equation in A, B, C is true, compared with FLT. ${\int_{0}^{A}{3 x^2 dx }}= {\int_{0}^{C-B}{3 (x+B)^2 dx }}$ Or: $A^3= {\int_{0}^{C-B}{3 (x^2+2Bx+B^2) dx }}$ Or (but we are for sure in R ): $A^3= (C-B)^3+3B(C-B)^2+3B^2(C-B))$ So after some tedious work in the infimus machine that magically makes mice=cats: $A^3= C^3-B^3$ And nothing change rising n... so we close the proof (I hope...) SO THE INFINITE DESCENT IS THE CONDITION TO OBTAIN A RESULT (if and only if) (I already show why for n=2 it works...) I hope CRGreathouse will digest and will say: ...this is the FLT proof... Thanks Ciao Stefano
May 18th, 2015, 12:16 PM   #2
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Quote:
 Originally Posted by complicatemodulus --> AT THIS POINT WE CAN SHOW BY COMPUTATION THAT RISING "K" WE NEVER FIND A RESULT OR WE ALWAYS HAVE BACK
More details?

May 18th, 2015, 10:16 PM   #3
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Quote:
 Originally Posted by CRGreathouse More details?
Thanks CRG, yes, this is the point I must clarify:

Computer helps us (not prove) in understanding the process I use, but I (hope) already show that this is very similar to the process we already know for "classic" integration :

- We start squaring a curve with wide columns, than we reduce step by step, showing that our approximation is in plus or minus

- As eight of the colums in classic integration we can keep the lower left corner height (or upper right)

- for a simple curve, with this kind of easy primitive function is easy to show that we get the result just when the base of the colums become infimus, so we all (now) believe in limits and in the integration process to square a curve.

- My integration process do the same: Mn over K represent a (little more) complicated squaring function, that becomes the primitive for K->infinity.

But there are 2 interval for x and gnomons: 0=<x<1 and 1<x<infinity

In 1<x<infinity our Gnomons function (X^n-(X-1)^n) perfectly square the curve... But to extend it in 0=<x<1 we need the shown trick...

I use as Height not the left or right corner, but the Gnomons value (X^n-(X-1)^n) that perfectly square X^n in the integers,

but, as shown also by graph in my old post, JUST when the step (or the base of the gnomons = 1/K) becomes littlest than 1, for what already shown by the knwon infinite descent proof,

the area of the new Mn/k Gnomons is always Littlest than the targhet area bellow the curve.

I already shown it by graph (was clear in mind, not in math, at that time), but Computer plot easy prove that with few sample: choose one case, for K=10 than K=100 etc.... and you can do that sampling n than n+1, nothing change (as knwon by Newton proof)

So for what I choose as Mn/K that become the primitive for K-> infinity, the area of this Gnomons is always undervalued from 1/k to an Integer P, respect to the area of X^n from 0 to P, still rising to very big value of K.

(Note that this process, exactly as known integration, can use the overextimated approximation of the curve, if we use the bigger gnomons. ...And this show what is also known as Dedekind section: we can rise K till what we want, our targhet is our of Q ).

The final point is: we are sure of the result C^n=A^n+B^n, since JUST for k->infinity the result appear, as shown right, from the integral.

And the point is exactly what already known (thanks to Newton first):

X^n curves from 1 to infinte can be easy squared by gnomons, but once we make the FLT trick we involves also the area bellow X^n in 0=< x <1, that needs infimus to be squared.

Hope no other stupid errors or bugs affect this.

Thanks
Ciao
Stefano

 May 18th, 2015, 10:17 PM #4 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry, forgotten (but already written here in several post), while for n=2 the primitive is a line, so can be squared with gnomons also in the 0-1 interval... So: FLT is right since for n>=3 the primitive is a curve. Much better (for our littlest mind) than 250 obscure pages with more obscure conclusion: "is right beacause all... are modular..." End of 500years long story ? Last edited by complicatemodulus; May 18th, 2015 at 10:24 PM.
 May 18th, 2015, 11:37 PM #5 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Still one little bug in, but just question of sings and regards the value of the step sum (step 1/k) in 1/k-1 (always 1), that is bigger than integral from 1/k to 1 that is 1 minus what before so from 0 to 1/k...
 May 19th, 2015, 12:19 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,622 Thanks: 2076 It's not over till the...
May 19th, 2015, 12:59 AM   #7
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Quote:
 Originally Posted by skipjack It's not over till the...
K (or your patience)-> infinity ;-P

May 19th, 2015, 04:26 AM   #8
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Quote:
 Originally Posted by skipjack It's not over till the...

 May 19th, 2015, 10:46 AM #9 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry typing error: the lower limit here is always 1/K like this $\sum_{x=1/K}^{A} {(3(x^2)/K -(3x)/K^2 +1/K^3)} =? \sum_{ x=1/K}^{C-B} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}$ --> AT THIS POINT WE CAN SHOW BY COMPUTATION THAT RISING "K" WE NEVER FIND A RESULT OR WE ALWAYS HAVE BACK : $\sum_{x=1/K}^{A} {(3(x^2)/K -(3x)/K^2 +1/K^3)} \neq \sum_{ x=1/K}^{C-B} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}$ AND WE ARE SURE FERMAT IS RIGHT SINCE IF, AND ONLY IF WE PASS TO THE LIMIT FOR K->INFINITY (SO VARIABLE IN R), WE HAVE AGAIN THE EQUALTY IS TRUE: 3) Make the limit with K-> infinity then pass to the definite integral. $\lim_{k\to\infty} { \sum_{x=1/K}^{A} {(3(x^2)/K -(3x)/K^2 +1/K^3)} = ? \lim_{k\to\infty}{ \sum_{ x=1/K}^{C-B} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}$ So 1/K for k_infinity goes to 0...
 May 19th, 2015, 09:24 PM #10 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 There is more to say about (for example): $\sum_{x=1/K}^{1} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}= 3X^2-3X+1|(X=(B+1)$ or: $\sum_{x=1/K}^{1} {(3(x+B)^2)/K -3(x+B)/K^2 +1/K^3)}= 3(B+1)^2-3(B+1)+1$ That is true for any n>=2, B, K integers... (proof is simple passing to the limit for k-> infinity) ... no answer, no interest ?

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