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May 30th, 2013, 10:57 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\:Nb" />. 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)\:ND1\rightarrow\sentence\:3)N^2D^N1=P_{XY}1" /> (1) On the other hand sentence 2b) Z^NN^2T\rightarrow N^2Z^NXY" /> 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<N) So: and we have two subcases here: ?a). Then and so: Given that N can't divide a (or we would have a=0) and N can't divide q since q<N, then by sentence 5 what remains is: ?b). Here we have: and then and likewise we now yield: In a way similar to the one we used in subcase Ia), we end to: and the conclusion is the same: Of course and In the same way, we show Case ??. N=6q1. Considering the same Euclidean division we conclude in an analogous way that: and we split it in two subcases again ???1). a+b=c and just like before: and so: [latex]\rightarrow and so: etc like before: etc, or and then again we end to the same IIb). a+b=N+c In a very analogous way, we get the same conclusion We will prove now that N=6q+1. Proof: From the following three relations, one at most can be true: Indeed, if we really had, e.g.: then which is false Therefore we assume, e.g.: Now if N=6q1, by sentence 1a we would have which is false, because N doesn't divide either ab or a+b as we have shown Now we can show that . Indeed, the first two hold by assumption. If we also had then: and since N doesn't divide b, we come to N=3, which is rejected by assumption. Now we will prove the uniqueness of a triple (a,b,c), that is, given two members of the triple,say a & b,the third is uniquely defined So if (a,b,c) is a triple and there was a second triple (a,b,c') then obviously: (or vice versa) which leads to: false Now we carry the relations from a,b,c to X,Y,Z. sentence 3) N^2Z^{3N}+X^{3N}" /> 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=XY 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 N3XY and since N doesn't divide 3 we conclude (sentence 5) r\:NY" /> 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+(ab)^NANa+BN+b â?? N^2 (ab)^NANa+BN+b ???? N^2X^{N1}1 â?? N^2X^NX=(AN+a)^NANa â??(??????? ??) N^2 a^NANa ??? ????? N^2b^NbBN ??? ??? N^2 (ab)^Na^N+b^N ??? ??????? ?? ??? ???????????? ??? a,b,c ??? ???????,??? b=câ??Ncbâ??Na, ????? mathformasses ????????????: 6 ???????: ???? ??? 10, 2012 1:33 am 

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