My Math Forum FLT reduced to 2th degree equations

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 April 26th, 2015, 12:36 AM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 FLT reduced to 2th degree equations It's possible to reduce the FLT also for n>=3 to a 2th degree equation. For what already shown (and using a new trick on): 1) $A^n= \sum_{x=1}^{A} (x^n-(x-1)^n)$ 2) $B^n= \sum_{x=1}^{B} (x^n-(x-1)^n)$ 3) $C^n= \sum_{x=1}^{C} (x^n-(x-1)^n)$ So if 4) $C^n= A^n+B^n$ than 5) $C^n= 2A^n + \sum_{x=A+1}^{B} (x^n-(x-1)^n)$ OR 6) $C^n= 2B^n - \sum_{x=A+1}^{B} (x^n-(x-1)^n)$ So we can call: 7) $Delta= \sum_{x=A+1}^{B} (x^n-(x-1)^n)$ So Delta is the quantity to be added or removed to obtain C^n. But we also know that is possible to "linearize", for example if n=ODD and n>=3 then we can write: 1b $A^n= \sum_{x=1}^{A^{((n-1)/2)}} 2Ax-A$ 2b $B^n= \sum_{x=1}^{B^{((n-1)/2)}} 2Bx-B$ 3b $C^n= \sum_{x=1}^{C^{((n-1)/2)}} 2Cx-C$ Now expressing A^n and B^n as areas of the primitive that is a triangle, so limits of the step sum step 1/k for k->infinity 1b $A^n= \lim_{k->infty} \sum_{x=1/k}^{A^{((n-1)/2)}} 2Ax /k -A/K^2$ 1b $A^n= \int_{0}^{A^{((n-1)/2)}} 2Ax dx$ so A^n and B^n as integral of the primitive (area of triangles), 1c $A^n= (A^{((n-1)/2)})*(A^{((n+1)/2)})/2$ So we can put in relation 2 rectangles: 8a $2A^n= (A^{((n-1)/2)})*(A^{((n+1)/2)})$ and the same for B^n 8b $2B^n= (B^{((n-1)/2)})*(B^{((n+1)/2)})$ And have clear that FLT ask if there is an integer solution for a rectangle that has an area C^n that is between this 2 areas at the known condition: 5) $C^n= 2A^n + Delta$ and 6) $C^n= 2B^n - Delta$ Where Delta is here an area that must be added to the area 2A^n, or subtracted from 2B^n. So we have now clear that the rectangular area (that is C^n) upper right corner P3, will lies on the line that pass trough the 2 rectagle upper right corners P1,P2: $P1( A^{((n-1)/2)}} , A^{((n+1)/2)}) ; P2 ( B^{((n-1)/2)} , B^{((n+1)/2)})$ So P3 is at a certain distance from P1, or at certain x from A^{((n-1)/2)} and at certain distance y=A^{((n+1)/2)} + Tg(alpha) And that distance x has to satisfy the conditions that the upper (Delta), and lower (Delta) areas are equal. So since 2A^n and 2B^n are for sure integers, if FLT has an integer solution, than also Delta must be an integer area. So the 2th degree Equation that SOLVES ANY "n" Odd is (5) = (6) so: $C^n= 2A^n + Delta = 2B^n - Delta$ So 6) $A^n + Delta - B^n= 0$ where Delta is the area: 9) $(A^{((n-1)/2)} + x)*(x*Tg(alpha)) + x* (A^{((n+1)/2)})$ where $Tg(alpha)= [2*(B^{((n+1)/2)}-A^{((n+1)/2)}]/ [(B^{((n-1)/2)}-A^{((n-1)/2)}]$ But my question is: is this enought, using some rules of the 2th dergee equation (not Wiles) to prove there are not integer solutions ? Thanks Ciao Stefano Last edited by complicatemodulus; April 26th, 2015 at 12:41 AM.
 April 26th, 2015, 01:04 AM #2 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Here the picture (remember n>=3 and n= Odd) Last edited by complicatemodulus; April 26th, 2015 at 01:12 AM.
 April 26th, 2015, 09:10 PM #3 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 OK, time out... The correct answer is: is not necessary to make the calculation since, as shown here also with pictures, FLT imply an (infimus) condition that just squares can satisfy. And, as in my suspect from the beginning, the final proof pass trough a differential equations, but what I not suspect, is that this differential equation is so easy, and under our nose from the beginning. Thanks, Ciao Stefano Sorry for the too many wrong trick posted here.... but it was my trip...
April 27th, 2015, 06:05 AM   #4
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 Originally Posted by complicatemodulus 9) $(A^{((n-1)/2)} + x)*(x*Tg(alpha)) + x* (A^{((n+1)/2)})$ where $Tg(alpha)= [2*(B^{((n+1)/2)}-A^{((n+1)/2)}]/ [(B^{((n-1)/2)}-A^{((n-1)/2)}]$ But my question is: is this enought, using some rules of the 2th dergee equation (not Wiles) to prove there are not integer solutions ?
Of course not. You still have very large degree, you just moved it to other variables.

 April 27th, 2015, 07:14 AM #5 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Yes thanks, but you miss my previous answer. It seems to me that there is a differential equation that solve the FLT problem. Thansk Ciao Stefano
 May 2nd, 2015, 10:42 PM #6 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I hope this will be more clear: FLT ask when the system of 3 equations has a result in the integers: (1) $C^n= A^n + B^n$ (2) $C^n= (A+ K1)^n$ (3) $C^n= (B+ K2)^n$ I show that the (1) can be written in 2 ways: (4a) $C^n= 2A^n + Delta$ (4b) $C^n= 2B^n - Delta$ So coupling the (4a) and (2), (4b) and (3) we have: (5a) $2A^n + Delta= (A+ K1)^n = A^n + K1^n + MixA$ (5b) $2B^n - Delta= (B+ K2)^n = B^n + K2^n + MixB$ Where MixA and MixB are the two mixed products of the 2 binomial developes. So FLT state that: (5a) $+ Delta= (A+ K1)^n = -A^n + K1^n + MixA$ (5b) $- Delta= (B+ K2)^n = -B^n + K2^n + MixB$ or summing the two: $-A^n + K1^n + MixA + (-B^n + K2^n + MixB)= 0$ or: $A^n + B^n= K1^n + MixA + K2^n + MixB$ Now using again the right develope for C^n: (6a) $C^n=A^n + K1^n + MixA = K1^n + MixA + K2^n + MixB$ gives: $A^n= K2^n + MixB$ and: (6b) $C^n=B^n + K2^n + MixB = K2^n + MixB + K1^n + MixA$ gives: $B^n= K1^n + MixA$ Ok, all this to say that we have a system and certain conditions... I'm looking for a shortcut to avoid other more complicate concerning: I already found and show that this system has a solution for any value of A (or B, or C) for n=2. So is it possible that there will be 2 solution for A, one for n=2 and another for the same A for n >=3 ? Thanks ciao Stefano

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