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 May 30th, 2013, 11:57 AM #1 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 A new approach to Fermat's last theorem In this topic I will try to prove the first case of Fermat's last theorem (FLT1 as it is often written), using only high school maths, well known at Fermat's time, with one only exception. All numbers involved are positive integers and N is a prime. We'll need the following sentences: 1. Fermat's little theorem: is divisible by N. Or Corollary: 1a. 1b. So if and because then 2a. If N is an odd prime then we also have: and the + or - signs in the left side match the respective + or - signs in the right side 2b. If N is an odd prime, we also have 3. If N is an odd prime, we also have 3?. The reverse also holds: Every prime greater than 3 can be written in the form 6q�1. 5. r\:N|b" />. This holds for more than two factors a,b 6. If the product of two coprime numbers is the pth power of a number, then each of those coprime numbers is also the pth power of a number: 7. So the goal is to prove that if N divides none of X,Y,Z and N>2 then we can't have It is obvious that we will assume: Step 1. Basic relations between X,Y,Z,N Let We first show that Suppose this wasn't true. Then there would be a prime Q common in these two numbers: and if and as Q is a prime, we get j=Q and so but since we conclude that which is false, because X and Y are coprimes. So: and from (1) we see that is all "hidden" in N: and as N is a prime: which is false. Therefore: and since: by sentence 6 we end to: N,R)=1" /> From Fermat's little theorem: and as N doesn't divide R, by sentence 5 it follows sentence 2a)\:N|D-1\rightarrow\ sentence\:3)N^2|D^N-1=P_{XY}-1" /> (1) On the other hand sentence 2b) Z^N-N^2T\rightarrow N^2|Z^N-X-Y" /> Likewise Combining these last three equalities, we get: (2) Step 2 . Consider the three Euclidean divisions with the three (positive) remainders a,b,c are smaller than N. Then: r\:0." /> By sentence 4, if N>3 (the case N=3 has been proved long ago), we consider two cases Case I. N=6q+1 Let us assume the following Euclidean division (m In the end of step 1 we proved: and so: But,as we showed: In the same way we prove: Step 3. The assumption that ? doesn't divide X,Y or Z must be abandoned Let W=X-Y and so: (5) From relation (1): and because (see relation (2)): it follows that: (because of (5)) and eliminating denominators: (6) ???? N can't divide W, or (5) would yield N|3XY and since N doesn't divide 3 we conclude (sentence 5) r\:N|Y" /> which is false. Thus, (6) gives, according to sentence 5 again: (7) Now's the time to use the only sentence that was unknown in the 17th century. It was proved by Mirimanoff in 1909: In the first case of Fermat's last theorem,if: then: Using this and (7), we immediately get: ( But by Fermat's little theorem: otherwise we would have false, as we proved. Therefore ( gives: (??????? ??) N^2Q+(a-b)^N-AN-a+BN+b �?? N^2| (a-b)^N-AN-a+BN+b ???? N^2|X^{N-1}-1 �?? N^2|X^N-X=(AN+a)^N-AN-a �??(??????? ??) N^2| a^N-AN-a ??? ????? N^2|b^N-b-BN ??? ??? N^2| (a-b)^N-a^N+b^N ??? ??????? ?? ??? ???????????? ??? a,b,c ??? ???????,??? b=c�??N|c-b�??N|a, ????? mathformasses ????????????: 6 ???????: ???? ??? 10, 2012 1:33 am Tags approach, fermat, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bruno59 Number Theory 135 March 8th, 2015 11:37 AM mente oscura Number Theory 10 June 6th, 2013 09:33 PM McPogor Number Theory 15 May 31st, 2011 08:31 AM smslca Number Theory 4 September 14th, 2010 09:00 PM xfaisalx Number Theory 2 July 25th, 2010 05:59 AM

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