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Is anybody still trying to find a simpler proof for the last theorem, which Fermat might have conceived. Does Wiles proof tell _ Weather a simpler proof can exist or not.
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Re: Fermat's last theorem
Thu, January 1, 2009 - 11:51 AMI do NOT have a simpler proof, but 'in my bones', I know that such a proof exists.
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Re: Fermat's last theorem
Thu, January 1, 2009 - 12:24 PMNo, Wile's doesn't imply that there is a simpler proof. In fact, no theory can without actually doing so.
People, I think, will always try to find a simpler proof for it is mind boggling that could be right about some many things and yet be wrong about how to prove them (many of his theorems don't appear provable with the math known in his day). -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 11:56 AMI don't have the background to fathom Wiles' proof. Nevertheless we can have fun with lemmas that may ultimately lead to a simpler proof of FLT.
We know that all Pythagorean Triples contain a number that is divisible by 3. Assuming that Fermat Triples exist, can you prove that this is true for them as well? -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 12:10 PMLarry,
"We know that all Pythagorean Triples contain a number that is divisible by 3."
Can you site a proof of this please?
Thanks,
Troy
PS: the whole point of Wiles proof was to prove the no Fermat triples exist, so no you can't prove that they would have a number divisible by 3. -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 1:14 PMAbout the Pythagorean Triples. I don't need to cite a proof; it's intuitively obvious.
From a modulo 3 perspective, the square of any positive integer is either a 0 or a 1.
BWOC, suppose that a relatively prime Pythagorean Triple: a^2 + b^2 = c^2
exists, such that neither a nor b nor c is divisible by 3.
Then you'd end up with
1 + 1 congruent to 1 modulo 3 (if I'm using the correct terminology).
And that result does not make sense.
So the original assumption is false. -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 2:34 PM"I don't need to cite a proof; it's intuitively obvious. "
It isn't intuitively obvious which is why you provided a proof.
Although, I haven't identified a integer squared whose modulo 3 is 2, what is the proof that the modulo 3 of an integer squared must be 0 or 1?
I'm not asking to be a pain, I'm rusty on my number theory. -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 4:50 PMEvery number has one of the three forms:
3k
3k+1
3k+2
Square those mod 3 and see what you get.
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Re: Fermat's last theorem
Wed, January 7, 2009 - 5:48 PMActually your theorem that all integers squared modulo 3 are 0 or 1 is incorrect.
2^2 mod 3 = 2 or -1 -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 5:59 PMTroy wrote:
"2^2 mod 3 = 2 or -1"
I don't follow. 2^2 = 4. Divide by 3, and the remainder is 1.
Moreover if the integer that you square is not divisible by 3, the remainder will always be 1.
That's all I was saying. -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 6:28 PMSorry, I'm being stupid...
(3 k) mod 3 = 0
(3 k + 1 ) mod 3 = 1
(3 k + 2) mod 3 = 2
Now I'm forgetting, is
m*n mod p = (m mod p) * (n mod p) -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 6:43 PMBasically yes. However you may have to divide the product by p to get a remainder that's less than p, and looks prettier. If you aren't comfortable with this nomenclature, then go with the more familiar algebraic notation. -
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Re: Fermat's last theorem
Wed, January 7, 2009 - 6:59 PMIt's been 17 yrs since I've done number theory, it takes time to remember but that doesn't mean discomfort.
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Re: Fermat's last theorem
Wed, January 7, 2009 - 7:12 PMTroy wrote:
"PS: the whole point of Wiles proof was to prove the no Fermat triples exist, so no you can't prove that they would have a number divisible by 3."
By having fun with lemmas, I meant that if we can prove 2 contradictory lemmas about Fermat Triples, then they are both vacuously true, and FTs do not exist. Voilà! A new proof for FLT. I think that that would qualify as fun.
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