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McPgr's Profile
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In Topic: Fermat's Last Theorem
22 May 2011 - 04:07 PM
Hello everybody.
The discussion on this Proof died and I'm not going to resurrect it.
But I think that found another way to prove FLT. The new version of the proof is based on the same expressions for a, b, c deduced in the previous one but is much shorter and in my opinion more consistent and easier to deal with.
The URL of the new version
To open with MS Office 2010
My link
To open with MS Internet Explorer
My link
The changes start with Lemma-9
Thanks to all interested. -
In Topic: Fermat's Last Theorem
03 November 2010 - 06:30 PM
[ name='phillip1882' timestamp='1288810584' post='301750']
Quote
in lemma 5, you have the equation a= vp and b = wq
and state that...
a = uwv +v^n; b = uwv +w^n; c = uwv +v^n +w^n.
i'm not sure this is correct.
for example, let's take the simple 3,4,5 triangle. (n = 2)
Read the note after eq (13). It says that the latter does not work with n=2. The eq (26) is true for this case with u=1 and any w and v -
In Topic: Fermat's Last Theorem
01 November 2010 - 07:33 PM
[ name='jakuta' timestamp='1288580098' post='301686']
Quote
You cannot assume the truth of all parts of the theorem and then prove the theorem. That is vacuous implication.
By claiming, a + k = c and b + i = c, with k, i integer , you have already asserted there are three positive integers a, b, c
Don't understand. If the statement says that an + bn = cn is not true for integers I must assume that it is true for integers and then find a contradiction compromising such assumtion. What can be proved by assumption that the equation is true for real decimal numbers? Only that the assumption is true.
Quote
.So, both k and i must be > 0. For the same reason, a, b, c > 0 or you arrive at a contradiction
No objections
Quote
But, then you assume the theorem you are trying to prove.
Not at all. The assumption is contrary to the statement of the theorem (see above). -
In Topic: Fermat's Last Theorem
28 October 2010 - 06:18 PM
name='Qfwfq' timestamp='1288264455' post='301638']
Quote
you also specify n will be considered prime; this explains why the first lemma might be useful but are you sure it doesn't lose generality?
Equation (74) deals with it.
Quote
But, of course, if n is prime then it must also be coprime with any other integer, including the two as per Version A, so the premise of prime n restricts you to that anyway. Actually, looking back at the premise of Version A it hits me I can't recall it being exploited; did I miss something? Can you sort it all out a little bit?
Since in version A n divides neither f nor k and according to lemma-2 is not divisor in (6a) and (6b) it plays no role in this version (except some features not used here). -
In Topic: Fermat's Last Theorem
28 October 2010 - 06:11 PM
[ name='Qfwfq' timestamp='1288264455' post='301638']
,Quote
you also specify n will be considered prime; this explains why the first lemma might be useful but are you sure it doesn't lose generality?
The eq (74) deals with it.
Quote
the premise of prime n restricts you to that anyway. Actually, looking back at the premise of Version A it hits me I can't recall it being exploited; did I miss something? Can you sort it all out a little bit?
Since in version A n divides neither f nor k and according to lemma-2 is not divisor in (6a) and (6b) it plays no role in this version (except some specific case not used here)
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