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

1. ## 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.

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

Lol. Welcome to the club!

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

Originally Posted by astrotech
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.
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..)

4. ## 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

5. ## 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 things

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

Originally Posted by Icarus496
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 things
The Binomial theorem expresses this, which is my proof, where iff or for Fermat's expression. Since rem(a.b.n) is not equal to zero (is a positive integer for n>2), Fermat's theorem is proved.

Look up the Binomial Expansion.. in Wiki...

7. ## 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.

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

Originally Posted by Icarus496
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.

It only reduces to c = a+b if a=0 or b= 0, in which case c = b or c=a, respectively.

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

Originally Posted by BuleriaChk

It only reduces to c = a+b if a=0 or b= 0, in which case c = b or c=a, respectively.
and all cases has nothing to do with FLT

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

Originally Posted by BuleriaChk

It only reduces to c = a+b if a=0 or b= 0, in which case c = b or c=a, respectively.
Let c=2 and a=b=1.

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

Page 1 of 21 12311 ... Last

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•