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February 25th, 2019, 12:56 AM  #1 
Newbie Joined: Feb 2019 From: Watertown NY USA Posts: 5 Thanks: 0  A Proof of Fermat's Conjecture
Hello, Welcoming myself I have a proof of Fermat's Conjecture at: https://www.dropbox.com/s/xxdtksxplf...29%20.pdf?dl=0 within my Dropbox folder at: https://www.dropbox.com/sh/gabop77dl...O0JOExzRa?dl=0 Also my Facebook page link at: https://www.facebook.com/retenshun Thanks for any interest I am Simon Roberts of Watertown NY U.S.A. 
February 25th, 2019, 03:22 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,926 Thanks: 2205 
How many previous attempted proofs have you devised? How many will you need to make before giving up?

February 25th, 2019, 04:45 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  The worst part is to think of all the ACTUAL mathematics these people could be doing with the same time and effort they expend on trying to prove FLT with elementary algebraic manipulations, disprove Cantor, or prove that $.999... \neq 1$, etc etc. Such a waste.

February 25th, 2019, 08:03 AM  #4 
Senior Member Joined: May 2013 Posts: 118 Thanks: 10 
Unlike skipjack and SDK, I will not judge before seeing. So I saw and I'm interested. Because your work, my dear Reyenshun, reminds me of mine. For a while, I've been up in the case pabc I used your x,y,z too, which I called remainders and I also proved your x+yz=0 (1), not by some "foggy" way, but by showing that p$\displaystyle ^2$x$\displaystyle ^{p1}$y$\displaystyle ^{p1}$ Eventually, I ended in x$\displaystyle ^p$+y$\displaystyle ^p$z$\displaystyle ^p$, which contradicts (1) of course. But I wouldn't post anything before addressing case pabc too. Now I see I might change my mind. As for your attempt, I've spotted at least one flaw, fatal or not I don't know. I'll have a second glance when I'll be able. Last edited by skipjack; March 12th, 2019 at 03:24 AM. 
February 25th, 2019, 07:53 PM  #5  
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
This is because any supposed proof based on algebraic manipulations (multiplication, addition, taking powers, inverses) would also hold equally well over the padics for p prime. But FLT has solutions over the padics for EVERY prime. Thus, there is no reason to go diving into his supposed proof to find the mistake. That doesn't mean I'm not confident it's there. Last edited by skipjack; March 12th, 2019 at 03:21 AM.  
March 11th, 2019, 05:07 PM  #6 
Newbie Joined: Feb 2019 From: Watertown NY USA Posts: 5 Thanks: 0  three or four, perhaps. 
March 11th, 2019, 05:12 PM  #7  
Newbie Joined: Feb 2019 From: Watertown NY USA Posts: 5 Thanks: 0  Quote:
Last edited by skipjack; March 12th, 2019 at 03:25 AM.  
March 14th, 2019, 02:57 AM  #8 
Senior Member Joined: May 2013 Posts: 118 Thanks: 10  Looks like I can't write down a sentence without a typo I meant that $\displaystyle x^p+y^pz^p=0$,which I ended to, contradicts x+yz=0 As for your OP: Ignoring for now that there is no explicit proof of x+y+z=0, can you please enlighten me about how (case $\displaystyle p\nmid abc$): $\displaystyle (x^p+y^p)/(x+y)$=$\displaystyle \prod_{r=1}^u(pQ_r+1)^{p1}$ Specifically how exponent p1 is dervied 

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conjecture, fermat, proof 
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