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 December 19th, 2014, 02:21 AM #1 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 Fermat the last... Knowing FLT conditions , so A
December 19th, 2014, 07:34 AM   #2
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Quote:
 Originally Posted by complicatemodulus ...No way to find a triplet A, B,C under Fermat conditions...
Everything else in your post aside, you don't demonstrate this.

 December 19th, 2014, 07:46 AM #3 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 Sorry I'm not Einstein... It's just on the way for... you probably can check, correct, and perhaps conclude. Thanks Ciao Stefano
 December 19th, 2014, 08:06 AM #4 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 With n=3, C=B+1 seems doesen't works, so no integer B satisfy the identity. I hope the rest will came... Thansk Ciao Stefano
 December 22nd, 2014, 07:44 AM #5 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 Here the actual revision on the Sum / Step sum / n-th problems: http://www.maruelli.com/COMPLICATE%2...tion%20def.pdf There are still many errors but I hope sum interesting point on. Thanks Ciao Stefano
 December 24th, 2014, 09:58 AM #6 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 Same link, I correct this wrong graph: Thanks & Merry Christmass ! Ciao Stefano
 December 27th, 2014, 09:10 AM #7 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 If I've not made some mistake in the develope: $\displaystyle -3{C^2} B+3CB^2 + {(3/2)}BC^2 +{(3/2)}B +(CB^2 - 3BC + C - B^3 - 3B^2+B) =0$ Has No integers solutions. -3*C^2*B+3*C*B^2+(3/2)*B*C^2+(3/2)*B+C*B^2-3*C*B+C-B^3-3*B^2+B=0 integer solutions - Wolfram|Alpha And the same into the rationals: -3*C^2*B+3*C*B^2+(3/2)*B*C^2+(3/2)*B+C*B^2-3*C*B+C-B^3-3*B^2+B=0 rational solutions - Wolfram|Alpha Thanks Ciao Stefano
 December 27th, 2014, 11:33 AM #8 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Of course W|A knows that FLT has already been solved by Wiles, and uses this to answer Diophantine queries.
 December 27th, 2014, 11:25 PM #9 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 Yes, always a round circle... I'm working on the differential equation, but is hard for me... Thanks Ciao Stefano
 March 4th, 2015, 06:14 AM #10 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 I forgot to write this to clarify my complicate-modulus conclusion on the FLT: Knowing FLT conditions n>=3 odd, so A3): (1) $A^3= \sum_{m =1}^{A} {(3m^2 -3m +1) } =? \sum_{X =B+1}^{C} {(3X^2 -3X +1)}$ But I know, also, that: $A^3 -A+ 1= integer$ And this is equal to: $A^3 -A+ 1= \sum_{m =1}^{A} {(3m^2 -3m +1/A) }$ So once you shift the right hand sum (1) to lower limit =1 (as already posted here) So you can apply the same division on the last term. The result must return an integer, if the sum is equal to the cube A^3. Than FLT follows for parity check on A,B,C on the divisibility of the new upper limit by A, in both case A,C ODDs, or A,C Evens. Thanks Ciao Stefano Last edited by skipjack; March 5th, 2015 at 11:52 AM.

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