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- 02-19-2017, 01:28 AM #1

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## Fermat's last, and mine too.

Fermat's was a note in the margin of a book claiming he had devised a simple proof that a^n+b^n=/=c^n for n>2.

Here's a simple proof.

when a=b, a^n+b^n<c^n for n>2. As a approaches 0, b approaches c. Therefore as a approaches 0, a^n+b^n approaches c^n for n>2.

End of story.Lies have the stench of death and defeat.

- 02-19-2017, 02:31 AM #2
## Re: Fermat's last, and mine too.

Lol. Welcome to the club!

- 02-21-2017, 01:30 PM #3

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## Re: Fermat's last, and mine too.

The problem is in the term "approach" if a,b, and c are integers, which is a catchall phrase for infinitesimals (derivatives).

The Binomial Theorem states it exactly (as does my proof) for positive integers, n >2:

, where (so only if or

(so or , respectively.

But I'm happy to see that Grapes seems to understands your version, even though it doesn't involve integers (). (There is a derivative relation between the Binomial Theorem and GTR as I've just posted in the other thread, where the Jacobian of the metric tensor is characterized as consisting of integers (v'/c' invariant as an "acceleration" )

The (general) relativistic relation is where a is identified with c in the relation v/c =b/c, so if a goes to zero v=b goes to zero as well and there is no integer to begin with (c=0) with all elements identified with differentials rather than integers.

(i.e, there is no metric tensor characterizing stress-energy)

(there is no curvature because there is nothing in the Universe... not even a parking lot..)Last edited by BuleriaChk; 02-21-2017 at 01:47 PM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

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- 02-24-2017, 02:55 PM #4

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## Re: Fermat's Theorem, Revistited

**Proof of Fermat's Theorem**

If Fermat's expression were true (), , , and would be a Pythagorean Triple. They are not a Pythagorean Triple by the Binomial Theorem, where rem(a,b,n) > 0.

**Therefore,****, for a,b,c,n positive integers, n>2**

**QED**

Triangle.png_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 02-26-2017, 04:44 PM #5

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## Re: Fermat's last, and mine too.

Hey, buleria,

Apologies in advance if I don't format this right, I have very little experience with latex

I just wanted to point out that your proof doesn't really work.

You're just saying that:

c=a+b

c^n=(a+b)^n

And then setting (a+b)^n equal to a^n+b^n,

so of course a or b has to be zero.

All you're saying is (a+b)^n=a^n+b^n IFF a or b=0

What you want to prove is that c^n = a^n+b^b IFF a or b=0

Those are different thingsLast edited by Icarus496; 02-26-2017 at 04:49 PM.

- 02-26-2017, 04:52 PM #6

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## Re: Fermat's last, and mine too.

Last edited by BuleriaChk; 02-26-2017 at 04:56 PM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 02-26-2017, 05:08 PM #7

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## Re: Fermat's last, and mine too.

Right, I understand the binomial expansion, but you can only apply the binomial theorem if you have something of the form (a+b)^n. If you set that equal to c^n, like you have to do for your proof, that reduces to c=a+b, which is not what you want to prove.

- 02-26-2017, 10:55 PM #8

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## Re: Fermat's last, and mine too.

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 02-27-2017, 02:28 AM #9

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- 02-27-2017, 03:05 AM #10

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## Re: Fermat's last, and mine too.

Let c=2 and a=b=1.

If you read the Wiki page you cited (under "Theorem statement"), it says , not for any c.

Added Link for user:

Cited Wikipedia PageLast edited by Neverfly; 02-27-2017 at 03:22 AM.

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