May 13th, 2017, 01:26 PM  #11 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,775 Thanks: 2194 Math Focus: Mainly analysis and algebra 
Or that, later in life, Fermat did write down a proof for a special case ($n=4$?). If he did indeed have a proof for all $n$, he'd have given it then.

May 13th, 2017, 08:20 PM  #12  
Senior Member Joined: Mar 2017 From: . Posts: 201 Thanks: 2 Math Focus: Number theory 
I will first put down what I have before finding a good algebra book. I might not find the point below in an algebra book Quote:
Proving such a theorem in different ways wouldn't rely much on what we already know but on what we can come up with. Last edited by Mariga; May 13th, 2017 at 08:24 PM.  
May 13th, 2017, 10:31 PM  #13 
Senior Member Joined: Mar 2017 From: . Posts: 201 Thanks: 2 Math Focus: Number theory 
Guys, It might be Fermat did it this way. Proving the theorem this way will not only prove the theorem itself but might be the proof that Fermat had the proof after all.

May 13th, 2017, 10:52 PM  #14 
Senior Member Joined: Aug 2012 Posts: 1,366 Thanks: 321  It might. But what are the odds that you know something the smartest mathematicians in the world haven't found in 350 years of looking?

May 13th, 2017, 11:18 PM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,775 Thanks: 2194 Math Focus: Mainly analysis and algebra 
Except that Fermat didn't prove it any more than you have. (Although he probably had a rather better idea).

May 14th, 2017, 01:32 AM  #16 
Senior Member Joined: Mar 2017 From: . Posts: 201 Thanks: 2 Math Focus: Number theory 
I've just said that because I am encountering some of his other related works in my research. The odds? I don't know.. Same as what happened with Collatz theorem.
Last edited by Mariga; May 14th, 2017 at 01:35 AM. 
May 14th, 2017, 04:07 AM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,775 Thanks: 2194 Math Focus: Mainly analysis and algebra  
May 14th, 2017, 10:28 PM  #18 
Senior Member Joined: Dec 2012 Posts: 951 Thanks: 23 
Still can't believe the focus was not on: Given A,B Integer and $\displaystyle A^n=\sum_{1}^{A}(X^n(X1)^n))$ $\displaystyle B^n=\sum_{1}^{B}(X^n(X1)^n))$ So Fermat can be rewritten as: $\displaystyle C^n = 2A^n+\Delta$ & $\displaystyle C^n = 2B^n\Delta$ with: $\displaystyle \Delta=\sum_{A+1}^{B}(X^n(X1)^n))$ $C^n$ is the unknown value is better to call $P$ or $Y_f$ since is an Ordinate on a cartesian plane that define what we can call the Fermat's point on a curve $Y=2X^n$ Due to Fermat's conditions the resulting $X_f$ is always an irrational due to the $2^{1/n}$ factor. $X_f$ is also the right border of the area bellow the derivate, and this area is equal to what we call $C^n$. In case $n=2$ the area bellow the derivate is a triangle, so it can be squared in the integers still if $X=F(sqr(2))$ while In case $n>2$ the First Derivate is a Curve So there is no way to square this derivate except going infimus. (here a simple example n=2 A=5, B=6) The proof is done with some simple concerning on where the $X_f$ is fiwed by Fermat's condition respect to the Medium Point on the line that join $(A,A^n)$; $(B, B^n)$ and the same on the derivate.... With a simple Xls table you can plot and see that $X_f$ is always too big respect to the Abscissa will have an integer (and a Rational too) Area corresponding to an Integer Power. Still working on to make better pictures and tables. Really can't believe nobody want share as public and known this formulation... (and in case prove or disprove I'm right or not...). Last edited by complicatemodulus; May 14th, 2017 at 10:59 PM. 
May 15th, 2017, 01:17 AM  #19 
Senior Member Joined: Dec 2012 Posts: 951 Thanks: 23 
Here the reason why Fermat is Right: Taking s example several Couple A and B integers:  Whith his trick he fix $Y'_f > Y_{MP}$ Or more simple he fix:  Whith his trick he fix $X_f > X_{BP}$ Where the BP is the Balancing Point for the Gnomon's Roof of B Rising the value $(BA)$, the difference rise, so we can be sure it doesen't need other check. So the first part of the Final Proof state that $P' (X_f,Y'f)$ is always more right than necessary, independently of who A and B are. I know looks like brute force, but here we can all see and say something of "math" and numbers.... 
May 15th, 2017, 10:52 PM  #20 
Senior Member Joined: Mar 2017 From: . Posts: 201 Thanks: 2 Math Focus: Number theory 
I've just read his proof for n=4 and I'll say he didn't prove it the way I am doing it. Though still if he had a proof it would still be algebraic. For this one, I know very well that any polynomial eg, $$a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6$$ cannot be factored into 6 factors, all integers, using $a$, $b$, or both, where $a$ and $b$ can be any integers except 0 for both and 1 for one of them. It can be factored as $$(a+b)^6$$ only if it has binomial coefficients. Coming up with a rigorous proof of such a claim is not easy but I am quite sure there exists a very beautiful proof for it that would make it the simplest proof of this theorem. Last edited by Mariga; May 15th, 2017 at 10:58 PM. 

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