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 June 13th, 2015, 01:25 AM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 FLT as parametric equation FLT for n=3 is equal to say: $(-2 B^3 k^3+3 B^2 k^3-3 B^2 k^2-B k^3+3 B k^2-2 B k+2 C^3 k^3+3 C^2 k^3-3 C^2 k^2+C k^3-3 C k^2+2 C k) - (2 A k-3 A k^2-3 A^2 k^2+A k^3+3 A^2 k^3+2 A^3 k^3)<> 0$ for A,B,C,k INTEGERS It comes from my step sum: $\sum_{X=1/k}^{A}{( 3X^2/k -3X/k^2+1/k^3)} = \sum_{X=B+1/k}^{C}{( 3X^2/k-3X/k^2+1/k^3)}$ that has always a result in R going to the limits for k->infinity Some one can help ? Wolframalpha doesen't... Thanks
 June 13th, 2015, 09:32 PM #2 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I'm very sorry I forgot to present FLT in this more easy (final) way: Asking if A^n =C^-B^n is equal to ask if (example for n=3): $\sum_{x=1}^{A}{( 3x^2 -3x+1)} = \sum_{x=B+1/k}^{C}{( 3x^2 -3x+1)}$ Or using, as shown in my old posts, my step sum is equal to ask if: putting X= k*x $\sum_{X=1/k}^{A}{( 3X^2/k -3X/k^2+1/k^3)} = \sum_{X=B+1/k}^{C}{( 3X^2/k-3X/k^2+1/k^3)}$ That is easy to prove HAS NO INTEGER SOLUTION since we can prove by computation that the right sum become more close to the left hand sum if werise k. So we can perform now the INFINITE DESCENT: we can RISE k trying to vanish the difference between this 2 terms but we can see ( by computation) that we are not able to do so, till we go to the limit for k-> infinity $lim_{{k\to\infty}} ( \sum_{X=1/k}^{A}{( 3X^2/k -3X/k^2+1/k^3)}) = lim_{{k\to\infty}} (\sum_{X=B+1/k}^{C}{( 3X^2/k-3X/k^2+1/k^3)})$ So JUST now is possible to restore the identity in R (now coming back to the general form of the 2th binomial term): (1) A^n = $\displaystyle \int_{1}^{A}{ [n*x^{(n-1)}]} dx = \int_{B}^{C}{ [ n(x)^{(n-1)}]} dx = C^n-B^n$ So we have Never a result with k in N or Q. So is not possible to perfectly fit the area of A^n, till k->infinity. I hope this was right and be (or not) the Infinte descent claimed by Fermat. What is nice is that this is very simple proof (in case right), and works for any n, since the inthernal term of the sums: Mn= (x^n-(x-1)^n) doesen't change the problem (except for n=2 where I already show what happen to the sum). Also I hope is clear why in the past I call FLT a collection of Dedekind Sections. For all the same reasons (I found on the way of 7 years long work) this solves also Beal: A,B,C non coprimes allow to find a K that divide all them, and x,y,z insthead of n, allow to find an integer solution. I show by graph that this solution is a solution of a linearized system of 3 lines, so there are few possible case for the soution. For sure a more math solution must be written, and I'm working on. Still kindly waiting for some good opinion about... (and an AMS answer in case...) Thanks Ciao Stefano Last edited by complicatemodulus; June 13th, 2015 at 10:01 PM.
 June 14th, 2015, 04:58 AM #3 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry here removing some refuse (1/k in the first sum): ************************************************** ** Asking if A^n =C^-B^n is equal to ask if (example for n=3): $\sum_{x=1}^{A}{( 3x^2 -3x+1)} = \sum_{x=B+1}^{C}{( 3x^2 -3x+1)}$ Or using, as shown in my old posts, my step sum is equal to ask if: putting X= k*x $\sum_{X=1/k}^{A}{( 3X^2/k -3X/k^2+1/k^3)} = \sum_{X=B+1/k}^{C}{( 3X^2/k-3X/k^2+1/k^3)}$ That is easy to prove HAS NO INTEGER SOLUTION since we can prove by computation that the right sum become more close to the left hand sum if werise k. So we can perform now the INFINITE DESCENT: we can RISE k trying to vanish the difference between this 2 terms but we can see ( by computation) that we are not able to do so, till we go to the limit for k-> infinity $lim_{{k\to\infty}} ( \sum_{X=1/k}^{A}{( 3X^2/k -3X/k^2+1/k^3)}) = lim_{{k\to\infty}} (\sum_{X=B+1/k}^{C}{( 3X^2/k-3X/k^2+1/k^3)})$ So JUST now is possible to restore the identity in R (now coming back to the general form of the 2th binomial term): (1) A^n = $\displaystyle \int_{1}^{A}{ [n*x^{(n-1)}]} dx = \int_{B}^{C}{ [ n(x)^{(n-1)}]} dx = C^n-B^n$ So we have Never a result with k in N or Q. So is not possible to perfectly fit the area of A^n, till k->infinity. I hope this was right and be (or not) the Infinte descent claimed by Fermat. What is nice is that this is very simple proof (in case right), and works for any n, since the inthernal term of the sums: Mn= (x^n-(x-1)^n) doesen't change the problem (except for n=2 where I already show what happen to the sum). Also I hope is clear why in the past I call FLT a collection of Dedekind Sections. For all the same reasons (I found on the way of 7 years long work) this solves also Beal: A,B,C non coprimes allow to find a K that divide all them, and x,y,z insthead of n, allow to find an integer solution. I show by graph that this solution is a solution of a linearized system of 3 lines, so there are few possible case for the soution. For sure a more math solution must be written, and I'm working on. Still kindly waiting for some good opinion about... Thanks Ciao Stefano
June 14th, 2015, 08:42 AM   #4
Banned Camp

Joined: Dec 2012

Posts: 1,028
Thanks: 24

Quote:
 Originally Posted by complicatemodulus FLT for n=3 is equal to say: $(-2 B^3 k^3+3 B^2 k^3-3 B^2 k^2-B k^3+3 B k^2-2 B k+2 C^3 k^3+3 C^2 k^3-3 C^2 k^2+C k^3-3 C k^2+2 C k) - (2 A k-3 A k^2-3 A^2 k^2+A k^3+3 A^2 k^3+2 A^3 k^3)<> 0$ for A,B,C,k INTEGERS
I forgot to say that this is the reply of wolframalpha when you push in, each my single step sum, step 1/k.

It doesen't recognize the limit for k->infinity, so the integral.

 June 14th, 2015, 10:09 AM #5 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Still some typing error: (>) where new. ************************************************** ** Asking if > $A^n=C^n -B^n$ is equal to ask if (example for n=3): $\sum_{x=1}^{A}{( 3x^2 -3x+1)} = \sum_{x=B+1}^{C}{( 3x^2 -3x+1)}$ Or using, as shown in my old posts, my step sum is equal to ask if: putting X= k*x $\sum_{X=1/k}^{A}{( 3X^2/k -3X/k^2+1/k^3)} = \sum_{X=B+1/k}^{C}{( 3X^2/k-3X/k^2+1/k^3)}$ That is easy to prove HAS NO INTEGER SOLUTION since we can prove by computation that the right sum become more close to the left hand sum if werise k. So we can perform now the INFINITE DESCENT: we can RISE k trying to vanish the difference between this 2 terms but we can see ( by computation) that we are not able to do so, till we go to the limit for k-> infinity $lim_{{k\to\infty}} ( \sum_{X=1/k}^{A}{( 3X^2/k -3X/k^2+1/k^3)}) = lim_{{k\to\infty}} (\sum_{X=B+1/k}^{C}{( 3X^2/k-3X/k^2+1/k^3)})$ So JUST now is possible to restore the identity in R (now coming back to the general form of the 2th binomial term): > (1) $A^n= \int_{0}^{A}{ [n*x^{(n-1)}]} dx = \int_{B}^{C}{ [ n(x)^{(n-1)}]} dx = C^n-B^n$ So we have Never a result with k in N or Q. So is not possible to perfectly fit the area of A^n, till k->infinity. I hope this was right and be (or not) the Infinte descent claimed by Fermat. What is nice is that this is very simple proof (in case right), and works for any n, since the inthernal term of the sums: $Mn= (x^n-(x-1)^n)$ doesen't change the problem (except for n=2 where I already show what happen to the sum). Also I hope is clear why in the past I call FLT a collection of Dedekind Sections. For all the same reasons (I found on the way of 7 years long work) this solves also Beal: A,B,C non coprimes allow to find a K that divide all them, and x,y,z insthead of n, allow to find an integer solution. I show by graph that this solution is a solution of a linearized system of 3 lines, so there are few possible case for the soution. For sure a more math solution must be written, and I'm working on. Still kindly waiting for some good opinion about... Thanks Ciao Stefano
 June 15th, 2015, 09:57 PM #6 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 To make a numerical example use: n=3 Using the sum of gnomons M3 or M/k to computate C: M3= 3x^2-3x+1 M3/k= 3x^2/k -3x/k^2 +1/k^3 A=7 ; A^3 = 343 B=10 ; B^3 = 1000 A^3+B^3 = 1343 C=? Using k=1 to try to find C we have: x - M3 - SUM - REST 1 1 1 2 7 8 3 19 27 4 37 64 5 61 125 6 91 216 7 127 343 8 169 512 9 217 729 10 271 1000 11 331 1331 12 12 397 1728 -385 At 11 we have rest 12, at 12 we overpass so rest is -385 So 11< C <12 Using K =7 we have: x - M3/7 - SUM - REST 1 0.142857143 0.002915452 0.002915452 2 0.285714286 0.020408163 0.023323615 3 0.428571429 0.055393586 0.078717201 4 0.571428571 0.10787172 0.186588921 5 0.714285714 0.177842566 0.364431487 6 0.857142857 0.265306122 0.629737609 7 1 0.370262391 1 8 1.142857143 0.49271137 1.49271137 9 1.285714286 0.632653061 2.125364431 10 1.428571429 0.790087464 2.915451895 11 1.571428571 0.965014577 3.880466472 12 1.714285714 1.157434402 5.037900875 13 1.857142857 1.367346939 6.405247813 14 2 1.594752187 8 15 2.142857143 1.839650146 9.839650146 16 2.285714286 2.102040816 11.94169096 17 2.428571429 2.381924198 14.32361516 18 2.571428571 2.679300292 17.00291545 19 2.714285714 2.994169096 19.99708455 20 2.857142857 3.326530612 23.32361516 21 3 3.67638484 27 22 3.142857143 4.043731778 31.04373178 23 3.285714286 4.428571429 35.47230321 24 3.428571429 4.83090379 40.303207 25 3.571428571 5.250728863 45.55393586 26 3.714285714 5.688046647 51.24198251 27 3.857142857 6.142857143 57.38483965 28 4 6.61516035 64 29 4.142857143 7.104956268 71.10495627 30 4.285714286 7.612244898 78.71720117 31 4.428571429 8.137026239 86.85422741 32 4.571428571 8.679300292 95.5335277 33 4.714285714 9.239067055 104.7725948 34 4.857142857 9.816326531 114.5889213 35 5 10.41107872 125 36 5.142857143 11.02332362 136.0233236 37 5.285714286 11.65306122 147.6763848 38 5.428571429 12.30029155 159.9766764 39 5.571428571 12.96501458 172.941691 40 5.714285714 13.64723032 186.5889213 41 5.857142857 14.34693878 200.9358601 42 6 15.06413994 216 43 6.142857143 15.79883382 231.7988338 44 6.285714286 16.55102041 248.3498542 45 6.428571429 17.32069971 265.6705539 46 6.571428571 18.10787172 283.7784257 47 6.714285714 18.91253644 302.6909621 48 6.857142857 19.73469388 322.425656 49 7 20.57434402 343 50 7.142857143 21.43148688 364.4314869 51 7.285714286 22.30612245 386.7376093 52 7.428571429 23.19825073 409.9358601 53 7.571428571 24.10787172 434.0437318 54 7.714285714 25.03498542 459.0787172 55 7.857142857 25.97959184 485.058309 56 8 26.94169096 512 57 8.142857143 27.9212828 539.9212828 58 8.285714286 28.91836735 568.8396501 59 8.428571429 29.93294461 598.7725948 60 8.571428571 30.96501458 629.7376093 61 8.714285714 32.01457726 661.7521866 62 8.857142857 33.08163265 694.8338192 63 9 34.16618076 729 64 9.142857143 35.26822157 764.2682216 65 9.285714286 36.3877551 800.6559767 66 9.428571429 37.52478134 838.180758 67 9.571428571 38.67930029 876.8600583 68 9.714285714 39.85131195 916.7113703 69 9.857142857 41.04081633 957.7521866 70 10 42.24781341 1000 71 10.14285714 43.47230321 1043.472303 72 10.28571429 44.71428571 1088.186589 73 10.42857143 45.97376093 1134.16035 74 10.57142857 47.25072886 1181.411079 75 10.71428571 48.5451895 1229.956268 76 10.85714286 49.85714286 1279.813411 77 11 51.18658892 1331 12 78 11.14285714 52.5335277 1383.533528 -40.5335277 At 11 we have rest 12, at 11.14 we overpass so rest is -40.533... So 11 < C < 11.14 With k=100 (last few lines) M3/100 x - M3/100 - Sum - REST 1100 11 3.626701 1331 12 1101 11.01 3.633301 1334.633301 8.366699 1102 11.02 3.639907 1338.273208 4.726792 1103 11.03 3.646519 1341.919727 1.080273 1104 11.04 3.653137 1345.572864 -2.572864 At 11.03 we have rest 1.080273, at 11.04 we overpass so rest is -2.572864... So 11.03 < C < 11.04 And so on till k= infinity where we know by math we have the solution. Thanks Ciao Stefano
 June 28th, 2015, 09:21 PM #7 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 What follow is worng (so NOT equeal to FLT) since comese from an error of Wolframalpha: $(-2 B^3 k^3+3 B^2 k^3-3 B^2 k^2-B k^3+3 B k^2-2 B k+2 C^3 k^3+3 C^2 k^3-3 C^2 k^2+C k^3-3 C k^2+2 C k) - (2 A k-3 A k^2-3 A^2 k^2+A k^3+3 A^2 k^3+2 A^3 k^3)<> 0$ If you try type in Wolframalpha the limit: $lim_{k\to\infty} \sum_{1/k}^{A} 3x^2/k-3x/k^2+1/k^3=$ The correct result is A^3, but it think: $lim_{k\to\infty} \sum_{1/k}^{A} 3x^2/k-3x/k^2+1/k^3= 0$ so it say (when): $(2 A k-3 A k^2-3 A^2 k^2+A k^3+3 A^2 k^3+2 A^3 k^3)$ More in my next post (still if nobody think there is somethink of good in it.

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