My Math Forum A new approach to Fermat's last theorem

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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: $a^N-a$ is divisible by N. Or $a^N=Nb+a$ Corollary: 1a. $N\dagger ab \rightarrow N| a^{N-1}-b^{N-1}$ 1b. So if $X^N+Y^N=Z^N$ and because $N|Z^N-Z,\:N|X^N-X\:and\:N|Y^N-Y$ then $N| X^N-X+Y^N-Y-Z^N+Z=-X-Y\rightarrow N| X+Y-Z=NV$ 2a. $N | (a_1+a_2+...+a_k) \leftrightarrow N | (a_1^N+a_2^N+...+a_k^N)$ If N is an odd prime then we also have: $N | (\pm a_1\pm a_2\pm...\pm a_k) \leftrightarrow N | (\pm a_1^N\pm a_2^N\pm...\pm a_k^N)$ and the + or - signs in the left side match the respective + or - signs in the right side 2b.$(N^ma+b)^N= AN^{m+1}+b^N$ If N is an odd prime, we also have $(N^ma-b)^N= AN^{m+1}-b^N$ 3. $N^m|a-b\rightarrow N^{m+1}|a^N-b^N$ If N is an odd prime, we also have $N^m|a+b\rightarrow N^{m+1}|a^N+b^N$ 3?. The reverse also holds: $N^{m+1}|a^N-b^N\:and\:N\dagger ab\rightarrow N^m|a-b$ $4.$ Every prime greater than 3 can be written in the form 6q±1. 5. $N|ab\rightarrow N|a\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: $(a,b)=1\:and\:ab=c^p\rightarrow a=d^p\:and\:b=e^p$ 7. $N|a^p\rightarrow N\a$ So the goal is to prove that if N divides none of X,Y,Z and N>2 then we can't have $X^N+Y^N=Z^N$ It is obvious that we will assume: $(X,Y)=(X,Z)=(Z,Y)=1$ Step 1. Basic relations between X,Y,Z,N Let $P_{xy}=\sum_{i=1}^{N}X^{N-i}Y^{i-1}$ We first show that $(X+Y,P_{XY})=1$ Suppose this wasn't true. Then there would be a prime Q common in these two numbers: $X+Y=GQ^g\:and\: HPQ^h=P_{XY}=\sum_{i=1}^{N}X^{N-i}Y^{i-1}=$$\sum_{i=1}^{N}X^{N-i}(GQ^g-X)^{i-1}=NX^{N-1}+GQ^gJ\rightarrow NX^{N-1}=UQ^u$ and if $(Q^u,X^{N-1})\neq1\rightarrow(Q,X)\neq1=j$ and as Q is a prime, we get j=Q and so $Q|X$ but since $Q|X+Y$ we conclude that $Q|Y$ which is false, because X and Y are coprimes. So: $(Q^u,X^{N-1})=1$ and from (1) we see that $Q^u$ is all "hidden" in N: $N=wQ^u$ and as N is a prime: $w=u=1\:and\: Q^u=Q=N\rightarrow N|X+Y\rightarrow (sentence Ib) N|Z$ which is false. Therefore: $(X+Y,P_{XY})=1$ and since: $(X+Y)P_{XY}=Z^N$ by sentence 6 we end to: $(X+Y)=R^N\:,\:P_{XY}=D^N,\:Z=RD\:and\N,R)=1" /> From Fermat's little theorem: $N|Z^N-Z\rihgtarrow N|R^ND^N-Z+Y+X-Y-X\rightarrow N|R^ND^N-NV-R^N\rightarrow N|R^ND^N-R^N=R^N(L^N-1)$ and as N doesn't divide R, by sentence 5 it follows $N|D^N-1\rightarrow\sentence 2a)\:N|D-1\rightarrow\sentence\:3)N^2|D^N-1=P_{XY}-1" /> (1) On the other hand $Z=RD=R(Nt+1)\rightarrow R=Z-NtR\:and X+Y=R^N=(Z-NtR)^N\rightarrow\sentence 2b) Z^N-N^2T\rightarrow N^2|Z^N-X-Y" /> Likewise $N^2|X^N-Z+Y\:and N^2|Y^N-Z+X$ Combining these last three equalities, we get: $N^2|Z^N-Z,\:N^2|Y^N-Y,\:N^2|X^N-X\:,N^2|Z^{N-1}-1,\:N2|Y^{N-1}-1,\:N2|X^{N-1)-1\:and\:N2|X+Y-Z$ (2) Step 2 . $N^2|X^2+XY+Y^2$ Consider the three Euclidean divisions $X=AN+a,\: Y=BN+b,;| Z=CN+c$ with the three (positive) remainders a,b,c are smaller than N. Then: $X^N+Y^N=Z^N \leftrightarrow (AN+a)^N+(BN+b)^N = (CN+c)^N \rightarrow (sentence 2b) A'N^2+a^N+B#39;N^2+b^N =$$C'N^2+c^N\rightarrow N^2|a^N+b^N-c^N\rightarrow N|a^N+b^N-c^N \rightarrow (sentence 2a) N|a+b-c \rightarrow a+b-c=N\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 (mN$ false Now we carry the relations from a,b,c to X,Y,Z. $N|a^3+c^3=(X-NA)^3+(Z-NC)^3\rightarrow N|X^3+Z^3\rightarrow\sentence 3) N^2|Z^{3N}+X^{3N}" />$\rightarrow N^2|Z^{3(N-1)}Z^3+X^{3(N-1)}X^3$ In the end of step 1 we proved: $N^2|X^{N-1}-1,\: N^2|Z^{N-1}-1$ and so: $N^2|(N^2m+1)^3Z^3+(N^2d+1)^3X^3\rightarrow N^2|Z^3+X^3=(Z+X)(Z^2-ZX+X^2)$ But,as we showed: $N\dagger c+a= X+Z-(A+C)N\rightarrow N\dagger X+Z\rightarrow N^2|Z^2-ZX+X^2$ In the same way we prove: $N^2| Y^2+YX+X^2,\:N^2|Z^2-ZY+Y^2,\:N^2|X^3+Z^3,\:N2|Y^3+Z^3,\:N2|X^3-Y^3$ Step 3. The assumption that ? doesn't divide X,Y or Z must be abandoned Let W=X-Y and so: $N^2|X^2+XY+Y^2= (W+Y)^2+(W+Y)Y+Y^2 = 3Y^2+3WY+W^2 =3Y(Y+W)+W^2 = 3YX+W^2$ (5) From relation (1): $N^2| P_{XY}-1=X^{N-1}-X^{N-2}Y+...-XY^{N-2}+Y^{N-1}-1$ and because (see relation (2)): $N^2| Y^{N-1}-1$ it follows that: $N^2| X^{N-1}-X^{N-2}Y+...+X^2Y^{N-3}-XY^{N-2}=$ (because of (5)) $[(N^2M-W^2)/3Y]^{N-1}-[(N^2M-W^2)/3Y]^{N-2}Y+...+[(N^2M-W^2)/3Y]^2Y^{N-3}-(N^2M-W^2)Y^{N-2}/3Y$ and eliminating denominators: $N^2| (N^2M-W^2)^{N-1}-(N^2M-W^2)^{N-2}3Y^2 +... + (N^2M-W^2)^23^{N-3}Y^{2(N-3)} - (N^2M-W^2)3^{N-2}Y^{2(N-2)}$ $= (sentence\:2b)\:N^2M_1+W^{2(N-1)}-N^2M_2+W^{2(N-2)}3Y^2+...+N^2M_{N-2}+W^43^{N-3}Y^{2(N-3)}-N^2M_{N-1}+W^23^{N-2}Y^{2(N-2)} =$ $K^2L +W^2[(W^2)^{N-1}-(3Y^2)^{N-1}]/(W^2-3Y^2)\rightarrow N^2| W^2[(W^2)^{N-1}-(3Y^2)^{N-1}]$ (6) ???? N can't divide W, or (5) would yield N|3XY and since N doesn't divide 3 we conclude (sentence 5) $N|X\r\:N|Y" /> which is false. Thus, (6) gives, according to sentence 5 again: $N^2| (W^2)^{N-1}-(3Y^2)^{N-1}= W^{2(N-1)}-3^{N-1}(Y^{N-1})^2 =$$W^{2(N-1)}-3^{N-1}(N^2D+1)^2\rightarrow N^2| W^{2(N-1)}-3^{N-1}$ (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: $X^N+Y^N=Z^N$ then: $N^2|3^{N-1}-1$ Using this and (7), we immediately get: $N^2| W^{2(N-1)}-1= (W^{N-1}-1)(W^{N-1}+1)$ ( But by Fermat's little theorem: $N| W^{N-1}-1\rightarrow N\dagger W^{N-1}+1$ otherwise we would have $N| W^{N-1} ? (sentence\:7) N|W$ false, as we proved. Therefore ( gives: $N^2| W^{N-1}-1= (X-Y)^{N-1}-1\rightarrow N^2| (X-Y)^N-X+Y = (AN+a-BN-b)^N-AN-a+BN+b =$(??????? ??) N^2Q+(a-b)^N-AN-a+BN+b â?? N^2| (a-b)^N-AN-a+BN+b ???? N^2|X^{N-1}-1 â?? N^2|X^N-X=(AN+a)^N-AN-a â??(??????? ??) 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=câ??N|c-bâ??N|a, ????? mathformasses ????????????: 6 ???????: ???? ??? 10, 2012 1:33 am

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