May 13th, 2017, 05:24 AM  #1 
Senior Member Joined: Mar 2017 From: . Posts: 275 Thanks: 5 Math Focus: Number theory  Simple(st) Proof of Fermat's Last Theorem.
\begin{align*} a^2+b^2&=c^2\\ c^2a^2&=b^2\\ c^2a^2&=(ca)(c+a)\\ b^2&=(ca)(c+a) \end{align*} This is possible iff there exists integers $k$, $x_1$, and $x_2$ such that $ca=kx_1^2$ and $ca=kx_2^2$ to give} $$b^2=k^2x_1^2x_2^2$$ For $n>2$, \begin{align*} a^n+b^n&=c^n\\ c^na^n&=b^n\\ c^na^n&=(ca)(c^{n1}+c^{n2}a+c^{n3}a^2+\cdots +ca^{n2}+a^{n1})\\ b^n&=(ca)(c^{n1}+c^{n2}a+c^{n3}a^2+\cdots +ca^{n2}+a^{n1}) \end{align*} This would hold iff $$b^n=(kx_1^n)(kx_2^n)(kx_3^n)\cdots (kx_n^n)$$ So that $$b^n=k^nx_1^nx_2^nx_3^n\cdots x_n^n$$ However,$(c^{n1}+c^{n2}a+c^{n3}a^2+\cdots +ca^{n2}+a^{n1})$ doesn't have common like terms and hence cannot be factored into a product of $n1$ terms. Hence, $$b^n\neq(kx_1^n)(kx_2^n)(kx_3^n)\cdots (kx_n^n)$$ thus $$a^n+b^n\neq c^n$$ Last edited by Mariga; May 13th, 2017 at 05:45 AM. 
May 13th, 2017, 06:18 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,672 Thanks: 741 
Here we go again...

May 13th, 2017, 06:52 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 904 Thanks: 359 
I may be wrong, but does not the fundamental theory of algebra say that the polynomial of degree n  1 CAN BE FACTORED

May 13th, 2017, 07:18 AM  #4 
Senior Member Joined: Mar 2017 From: . Posts: 275 Thanks: 5 Math Focus: Number theory  
May 13th, 2017, 08:06 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,151 Thanks: 2387 Math Focus: Mainly analysis and algebra 
Where's that cartoon/blog about solving Fermat in the morning and Goldbach in the afternoon.

May 13th, 2017, 08:27 AM  #6 
Senior Member Joined: Mar 2017 From: . Posts: 275 Thanks: 5 Math Focus: Number theory 
Is there anything wrong with the proof?

May 13th, 2017, 09:29 AM  #7  
Senior Member Joined: May 2016 From: USA Posts: 904 Thanks: 359  Quote:
$\left ( c  \dfrac{\ a + ai\sqrt{3})}{2} \right ) \left ( c  \dfrac{\ a  ai\sqrt{3})}{2} \right) =$ $c^2  \dfrac{\ ac + aci\sqrt{3}}{2}  \dfrac{\ ac  aci\sqrt{3}}{2} + \dfrac{(\ a)^2  (ai\sqrt{3})^2}{4}=$ $c^2  \dfrac{\ 2ac }{2} + \dfrac{a^2  a^2(\ 1)(3)}{ 4} = c^2 + ac + \dfrac{a^2 + 3a^2}{4} = c^2 + ac + a^2.$ The polynomial CAN be factored, just not in the integers. So you have proved FLT for n = 3, but the objective is to prove it for any n over 2. You have not proved that the resulting polynomial in n  1 cannot be factored over the integers for all n over 2. You have asserted it cannot be factored, which is just wrong. You are not even close. Last edited by JeffM1; May 13th, 2017 at 09:35 AM.  
May 13th, 2017, 11:18 AM  #8 
Senior Member Joined: Mar 2017 From: . Posts: 275 Thanks: 5 Math Focus: Number theory 
I get your point. Though I don't think it is far from the solution. I already have a way around that. It can't be longer than two A4 pages. Unless I will have skipped some points just as I have. Is there any proof based on this method? Also I'd like to take a break from maths for a while. But still, I am open to collaborations. Last edited by Mariga; May 13th, 2017 at 11:34 AM. 
May 13th, 2017, 01:56 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,151 Thanks: 2387 Math Focus: Mainly analysis and algebra 
No, but there are loads of failed ones.

May 13th, 2017, 02:13 PM  #10 
Senior Member Joined: Aug 2012 Posts: 1,709 Thanks: 458 
The very nature of Wiles's proof provides empirical evidence that Fermat never had a proof nor does anyone not trained in research level algebraic geometry have a proof. Wiles's work lies in the domain of Grothendieck's revolution in algebraic geometry. People have often heard of Grothendieck, and Wiles's work lives within the world of that revolution. It didn't exist before the second half of the 20th century and it's simply so abstract that nobody in the 17th century could have anticipated it. Ditto for today's amateurs dabbling in high school algebraic proofs. Of course my argument does not make any logical case. There's no reason that Fermat didn't conceive of some special case of something very deep, and would have written it down but for the narrowness of the margin. I wasn't there inside Fermat's head. Nobody was. But as circumstantial evidence, you have really got to understand that Wiles's proof could not have existed before the categorical revolution in algebra of the post 1950s. It's very different than prior mathematics. And nobody's found a bridge that gets you from high school math to what the professionals are doing in the modern era. If OP has a proof for n = 3 (I'm no expert) that is very cool. I'd suggest reading a book on abstract algebra. That's how you get at FLT. Last edited by Maschke; May 13th, 2017 at 02:31 PM. 

Tags 
fermat, proof, simplest, theorem 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Fermat was correct... he had a truly marvelous simple proof of his last theorem.  RanberSingh  Number Theory  2  April 27th, 2017 09:06 AM 
Simple Proof of Fermat's Last Theorem and Beal's Conjecture  MrAwojobi  Number Theory  1  September 29th, 2015 08:22 AM 
Simple Fermat Proof a^n+b^n=c^n  M_B_S  Number Theory  15  July 29th, 2015 06:14 AM 
Analytical simple proof of Fermat's last theorem ????  RanberSingh  New Users  3  August 10th, 2014 02:44 PM 
Simple Proof of Fermat's Last Theorem and Beal's Conjecture  MrAwojobi  Number Theory  21  January 8th, 2011 10:11 AM 