My Math Forum Fermat the Last (I hope)

 Number Theory Number Theory Math Forum

 June 27th, 2016, 12:39 AM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Fermat the Last (I hope) Here the last turn for FLT: (for clarify what Step Sum, step $1/K$ are and works, pls check my several post on) Giving for true under Fermat's conditions (for $n>2$, $A2$ can be summarized as: it's due to mixed terms of higher degree (from 2). And a short cut will be: first derivate = curve. I know one or more bugs can affect this "process" I won't call "proof", so I'll wait for an expert check... (sub judice) It was a very hard work for me (8 years long)... I hope it's close to the full stop (for many reasons). Thanks Stefano
 June 27th, 2016, 10:45 AM #2 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry, still 2 computation bugs, but result don't change in the behaviour.... I'll check better asap... Thanks Stefano
 June 27th, 2016, 09:34 PM #3 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Here the last version with fixed develope bugs: On the right hand we will have (here first error fixed): $3B (C-B-1/K)^2 + 3B^2/K (C-B-1/K) = 3BC^2-3CB^2-6BC/K+3B^2/K+3B/K^2$ Reordering (here the second error fixed): $(A+B-C)^3 =$ $-3C(A+B-C)^2+3B(A+B-C)^2+3/K(A+B-C)^2-(-6ABC/K+3AB^2/K+3AC^2/K+9BC^2/K-9CB^2/K+ 3B^3/K-3C^3/K+6AB/K^2-6AC/K^2-12BC/K^2+6B^2/K^2+6C^2/K^2+3A/K^2+3B/K^2-3C/K^2)+ 3BC^2-3CB^2-6BC/K+3B^2/K+3B/K^2$ $= -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3+6ABC/K+12CB^2/K-3AB^2/K- 3AC^2/K-9BC^2/K+3C^3/K-6B^3/K+6AB/K-6AC/K-12BC/K+3A^2/K+3C^2/K+9B^2/K+6AC/K^2+ 12BC/K^2-6AB/K^2-6C^2/K^2-9B^2/K^2+3C/K^2-3A/K^2$ that if $K\to\infty$ becomes: $(A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3$ So it's clear that the right hand is NOT the right tri-nomial develop, so this is the absurdum we are looking for. Thanks Ciao Stefano
 June 29th, 2016, 11:17 PM #4 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 ...I hope someone kindly answer... The point is if it's possible to works with the limits for $K\to\infty$ in this way or not, so if the turn it's true, or false... I hope it's true since moving parts from left to right or viceversa we respect the inequality, just respecting the sing rule, and for the limits since we have just Sums nothing has to change... I already did another trip, just doble longer, where will be clear that assuming $A^=C^n-B^n$ for true in the integers there is no continuty if we assume, and computate in this way from right (+1/K) or from the left side (-1/K), since the two limits, once reduced in this way are different. But I still be prepared to the worst... Thanks Ciao Stefano
 July 7th, 2016, 12:36 AM #5 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 OK, I've a nice discussion with italian friends... (that just erase the topics ...) It's clear that: $(A+B-C)^3 = -12ABC+3BA^2+6AB^2+6AC^2+12BC^2-3CA^2-15CB^2+6B^3-3C^3$ is NOT the right tri-nomial develop, but it has, unfotunately, some integer solution... so what we have to do is to compare this result to the result coming from the limit from the right side so of: (3) $\displaystyle (A+1/K)^n= (C+1/K)^n-B^n$ At the moment it (once reduced with sum and pushed to the limit) give to me another result, so seems the two limits are different and this is the right absurdum since we know they must be the same on a continuos function... I will check the computation to avoid errors, and I've also to check witch of the possible cooling, or warming, is the shortest one to show the absurdum. ...it's a long way to tipperary...it's a long way to go... Thanks Ciao Stefano
 July 7th, 2016, 02:10 AM #6 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I check: (3) $\displaystyle (A-1/K)^n= (C)^n-(B-1/K)^n$ and sems to finish at: $\displaystyle (A+B-C)^3 = 3BC^2-6CB^2+3B^3$ So if all computation is right I'm alowd to say that this prove that the initial equation must be wrong ? Thanks Ciao Stefano

 Tags fermat, hope

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jeneferhokins New Users 9 July 12th, 2012 08:35 PM JOANA Abstract Algebra 1 May 5th, 2010 12:20 PM MathChallenged2010 New Users 7 February 25th, 2010 10:15 PM Rooonaldinho Algebra 1 April 7th, 2008 01:52 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top