December 19th, 2014, 01:21 AM  #1 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24  Fermat the last...
Knowing FLT conditions , so A<B<C integers Asking if C^B^n = A^n From the identity in R: (1) $\displaystyle \int_{B}^{C}{ [n*x^{(n1)}]} dx = \int_{0}^{CB}{ [ n(x+B)^{(n1)}]} dx = C^nB^n $ I've shown how to rewrite in N (see my previous posts) So we check if there is a solution to the (1) in N (in the example n=3): We know we can make and Exchange of variable calling $\displaystyle X=(x+B) $ $\displaystyle \sum_{m =B+1}^{C} {(3m^2 3m +1) } =? \sum_{X =1}^{CB} {(3X^2 3X +1)} $ or: $\displaystyle \sum_{m =B+1}^{C} {(3m^2 3m +1)} =? \sum_{x =1}^{CB} {(3(x+B)^2 3(x+B) +1)} $ or: $\displaystyle C^3B^3 =? \sum_{x =1}^{CB} {(3x^2 3x +1)} + 3B \sum_{x =1}^{CB} (x) + (CB)*(B^23B+1) $ or: $\displaystyle C^3B^3 =? (CB)^3 + 3B (C*(C+1)/2) + (CB^23BC+CB^33B^2+B) $ or: $\displaystyle A^3 =? C^3B^3 =? C^3B^33{C^2} B+3CB^2 + {(3/2)}BC^2 +{(3/2)}B +(CB^2  3BC + C  B^3  3B^2+B) $ etc..... The same can be done in Q using my step sum: $\displaystyle \sum_{x =B+1/k}^{C}(s) {[3x^2/k3x/k^2+1/k^3]} =? \sum_{x =1/k}^{CB} {(3(x+B)^2 /k 3(x+B)/k^2 +1/k^3)} $ ...No way to find a triplet A, B,C under Fermat conditions... Thanks Ciao Stefano Last edited by complicatemodulus; December 19th, 2014 at 01:59 AM. 
December 19th, 2014, 06:34 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
December 19th, 2014, 06:46 AM  #3 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24 
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, 07:06 AM  #4 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24 
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, 06:44 AM  #5 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24 
Here the actual revision on the Sum / Step sum / nth 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, 08:58 AM  #6 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24 
Same link, I correct this wrong graph: Thanks & Merry Christmass ! Ciao Stefano 
December 27th, 2014, 08:10 AM  #7 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24 
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^23*C*B+CB^33*B^2+B=0 integer solutions  WolframAlpha And the same into the rationals: 3*C^2*B+3*C*B^2+(3/2)*B*C^2+(3/2)*B+C*B^23*C*B+CB^33*B^2+B=0 rational solutions  WolframAlpha Thanks Ciao Stefano 
December 27th, 2014, 10:33 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Of course WA knows that FLT has already been solved by Wiles, and uses this to answer Diophantine queries.

December 27th, 2014, 10:25 PM  #9 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24 
Yes, always a round circle... I'm working on the differential equation, but is hard for me... Thanks Ciao Stefano 
March 4th, 2015, 05:14 AM  #10 
Senior Member Joined: Dec 2012 Posts: 1,014 Thanks: 24 
I forgot to write this to clarify my complicatemodulus conclusion on the FLT: Knowing FLT conditions n>=3 odd, so A<B<C integers Asking if Is equal to ask if (here the example n=3, but is the same if n>3): (1) But I know, also, that: And this is equal to: 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 10:52 AM. 

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