My Math Forum Simple(st) Proof of Fermat's Last Theorem.

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 May 13th, 2017, 04:24 AM #1 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Simple(st) Proof of Fermat's Last Theorem. \begin{align*} a^2+b^2&=c^2\\ c^2-a^2&=b^2\\ c^2-a^2&=(c-a)(c+a)\\ b^2&=(c-a)(c+a) \end{align*} This is possible iff there exists integers $k$, $x_1$, and $x_2$ such that $c-a=kx_1^2$ and $c-a=kx_2^2$ to give} $$b^2=k^2x_1^2x_2^2$$ For $n>2$, \begin{align*} a^n+b^n&=c^n\\ c^n-a^n&=b^n\\ c^n-a^n&=(c-a)(c^{n-1}+c^{n-2}a+c^{n-3}a^2+\cdots +ca^{n-2}+a^{n-1})\\ b^n&=(c-a)(c^{n-1}+c^{n-2}a+c^{n-3}a^2+\cdots +ca^{n-2}+a^{n-1}) \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,$(c^{n-1}+c^{n-2}a+c^{n-3}a^2+\cdots +ca^{n-2}+a^{n-1})$ doesn't have common like terms and hence cannot be factored into a product of $n-1$ 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 04:45 AM.
 May 13th, 2017, 05:18 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,413 Thanks: 1024 Here we go again...
 May 13th, 2017, 05:52 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 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, 06:18 AM   #4
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 Originally Posted by JeffM1 I may be wrong, but does not the fundamental theory of algebra say that the polynomial of degree n - 1 CAN BE FACTORED
Let's try this.

$$c^3-a^3=(c-a) (c^2+ca+a^2)$$

I'd like you to factor $(c^2+ca+a^2)$ into a product of two terms.

 May 13th, 2017, 07:06 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,656 Thanks: 2634 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, 07:27 AM #6 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory Is there anything wrong with the proof?
May 13th, 2017, 08:29 AM   #7
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 Originally Posted by Mariga Let's try this. c^3-a^3=(c-a) (c^2+ca+a^2)$I'd like you to factor$(c^2+ca+a^2)$into a product of two terms. The polynomials of degree n - 1 can all be factored. It is easy to give that factoring if n = 3.$\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 08:35 AM.

 May 13th, 2017, 10:18 AM #8 Banned Camp   Joined: Mar 2017 From: . Posts: 338 Thanks: 8 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 10:34 AM.
 May 13th, 2017, 12:56 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,656 Thanks: 2634 Math Focus: Mainly analysis and algebra No, but there are loads of failed ones.
 May 13th, 2017, 01:13 PM #10 Senior Member   Joined: Aug 2012 Posts: 2,306 Thanks: 706 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 01:31 PM.

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