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February 25th, 2019, 12:56 AM   #1
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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.
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February 25th, 2019, 03:22 AM   #2
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How many previous attempted proofs have you devised? How many will you need to make before giving up?
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February 25th, 2019, 04:45 AM   #3
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Quote:
Originally Posted by skipjack View Post
How many previous attempted proofs have you devised? How many will you need to make before giving up?
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.
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February 25th, 2019, 08:03 AM   #4
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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 p|abc
I used your x,y,z too, which I called remainders and I also proved your x+y-z=0 (1),
not by some "foggy" way, but by showing that
p$\displaystyle ^2$|x$\displaystyle ^{p-1}$-y$\displaystyle ^{p-1}$
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 p|abc 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.
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February 25th, 2019, 07:53 PM   #5
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Originally Posted by bruno59 View Post
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.
There is a simple theorem which proves that algebraic manipulation is incapable of producing a proof of FLT. Not "we haven't found one yet" or "maybe if I am smarter than anyone else who tried...", but rather, it's a THEOREM that it can't be done.

This is because any supposed proof based on algebraic manipulations (multiplication, addition, taking powers, inverses) would also hold equally well over the p-adics for p prime. But FLT has solutions over the p-adics 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.
Thanks from topsquark

Last edited by skipjack; March 12th, 2019 at 03:21 AM.
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March 11th, 2019, 05:07 PM   #6
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three or four, perhaps.
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March 11th, 2019, 05:12 PM   #7
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Quote:
Originally Posted by bruno59 View Post
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 p|abc
I used your x,y,z too, which I called remainders and I also proved your x+y-z=0 (1),
not by some "foggy" way, but by showing that
p$\displaystyle ^2$|x$\displaystyle ^{p-1}$-y$\displaystyle ^{p-1}$
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 p|abc 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.
How does x^p + y^p - z^p contradict x + y - z = 0?

Last edited by skipjack; March 12th, 2019 at 03:25 AM.
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March 14th, 2019, 02:57 AM   #8
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Quote:
Originally Posted by retenshun View Post
How does x^p + y^p - z^p contradict x + y - z = 0?
Looks like I can't write down a sentence without a typo
I meant that $\displaystyle x^p+y^p-z^p=0$,which I ended to, contradicts x+y-z=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 $\displaystyle p\nmid abc$):
$\displaystyle (x^p+y^p)/(x+y)$=$\displaystyle \prod_{r=1}^u(pQ_r+1)^{p-1}$
Specifically how exponent p-1 is dervied
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