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

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 May 30th, 2013, 10:32 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 class 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 6qï¿½1. 5. $N|ab\rightarrow N|a\:or\:N|b$. 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 was'nt 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}=$$\sum_{i=1}^{N}X^{N-i}(GQ^g-X)^{i-1}=NX^{N-1}+GQ^gJ\rightarrow NX^{N-1}=UQ^u$ and if $(Q^u,X^{N-1})\neq1\rightarrow(Q,X)\neq1=j$ and as Q is a prime,we get j=Q and so $Q|X$ but since $Q|X+Y$ we conclude that $Q|Y$ which is false,because X and Y are coprimes.So: $(Q^u,X^{N-1})=1$ and from (1) we see that $Q^u$ is all "hidden" in N: $N=wQ^u$ and as N is a prime: $w=u=1\:and\: Q^u=Q=N\rightarrow N|X+Y\rightarrow (sentence\:1b)\:N|Z$ which is false.Therefore: $(X+Y,P_{xy})=1$ and since: $(X+Y)P_{xy}=Z^N$ by sentence 6 we end to: $(X+Y)=R^N\:,\:P_{xy}=D^N,\:Z=RD\:and\:(N,R)=1$ From Fermat's little theorem: $N|Z^N-Z=R^ND^N-Z+Y+X-Y-X=(sentence\:1b)\:R^ND^N-NV-R^N\rightarrow N|R^ND^N-R^N=R^N(D^N-1)$ and as N doesn't divide R,by sentence 5 it follows: $N|D^N-1\rightarrow\:(sentence\:2a)\:N|D-1\rightarrow\:(sentence\:3)\:N^2|D^N-1=P_{xy}-1$ (1) On the other hand: $Z=RD=R(Nt+1)\rightarrow R=Z-NtR\:and X+Y=R^N=(Z-NtR)^N\rightarrow\:(sentence\:2b)\: Z^N-N^2T\rightarrow N^2|Z^N-X-Y$ Likewise $N^2|X^N-Z+Y\:and N^2|Y^N-Z+X$ Combining these last three equalities we get: $N^2|Z^N-Z,\:N^2|Y^N-Y,\:N^2|X^N-X,\:N^2|Z^{N-1}-1,\:N^2|Y^{N-1}-1,\:N^2|X^{N-1)-1\:and\:N^2|X+Y-Z$ (2) Step 2 . .$N^2|X^2+XY+Y^2$ Consider the three Euclidean divisions $X=AN+a,\: Y=BN+b,\: Z=CN+c$ with the three (positive) remainders a,b,c being smaller than N.Then: $X^N+Y^N=Z^N \leftrightarrow (AN+a)^N+(BN+b)^N = (CN+c)^N \rightarrow (sentence\:2b)\: A'N^2+a^N+B#39;N^2+b^N =$$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\:or\:0.$ In any case,it's easy to show that: $N^2|(a+b)^N-a^N-b^N,\:N^2|(a-c)^N-a^N+c^N,\:N^2|(b-c)^M-b^N+c^N$ (3) 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}$$\rightarrow N^2|Z^{3(N-1)}Z^3+X^{3(N-1)}X^3$ In the end of step 1 we proved: $N^2|X^{N-1}-1,\: N^2|Z^{N-1}-1$ and so: $N^2|(N^2m+1)^3Z^3+(N^2d+1)^3X^3\rightarrow N^2|Z^3+X^3=(Z+X)(Z^2-ZX+X^2)$ But,as we showed: $N\dagger c+a= X+Z-(A+C)N\rightarrow N\dagger X+Z\rightarrow N^2|Z^2-ZX+X^2$ In the same way we prove: $N^2| Y^2+YX+X^2,\:N^2|Z^2-ZY+Y^2,\:N^2|X^3+Z^3,\:N^2|Y^3+Z^3,\:N^2|X^3-Y^3$ Step 3. The assumption that ? does't divide X,Y or Z must be abandoned (quod erat demonstrandum) Let W=X-Y and so: $N^2|X^2+XY+Y^2= (W+Y)^2+(W+Y)Y+Y^2 = 3Y^2+3WY+W^2 =3Y(Y+W)+W^2 = 3YX+W^2$ (5) From relation (1): $N^2| P_{XY}-1=X^{N-1}-X^{N-2}Y+...-XY^{N-2}+Y^{N-1}-1$ and because (see relation (2)): $N^2| Y^{N-1}-1$ it follows that: $N^2| X^{N-1}-X^{N-2}Y+...+X^2Y^{N-3}-XY^{N-2}=\:(because\:of\:(5))$ $[(N^2M-W^2)/3Y]^{N-1}-[(N^2M-W^2)/3Y]^{N-2}Y+...+[(N^2M-W^2)/3Y]^2Y^{N-3}-(N^2M-W^2)Y^{N-2}/3Y$ and eleminating denominators: $N^2| (N^2M-W^2)^{N-1}-(N^2M-W^2)^{N-2}3Y^2 +... + (N^2M-W^2)^23^{N-3}Y^{2(N-3)} - (N^2M-W^2)3^{N-2}Y^{2(N-2)}$ $= (sentence\:2b)\:N^2M_1+W^{2(N-1)}-N^2M_2+W^{2(N-2)}3Y^2+...+N^2M_{N-2}+W^43^{N-3}Y^{2(N-3)}-N^2M_{N-1}+W^23^{N-2}Y^{2(N-2)} =$ $K^2L +W^2[(W^2)^{N-1}-(3Y^2)^{N-1}]/(W^2-3Y^2)\rightarrow N^2| W^2[(W^2)^{N-1}-(3Y^2)^{N-1}]$ (6) ???? N can't divide W, or (5) would yield N|3XY and since N doesn't divide 3 we conclude (sentence 5) $N|X\:or\:N|Y$ which is false.Thus,(6) gives,according to sentence 5 again: $N^2| (W^2)^{N-1}-(3Y^2)^{N-1}= W^{2(N-1)}-3^{N-1}(Y^{N-1})^2 =$$W^{2(N-1)}-3^{N-1}(N^2S+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 and it goes like this: 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: $N^2| W^{2(N-1)}-1= (W^{N-1}-1)(W^{N-1}+1)$ (8) But by Fermat's little theorem: $N| W^{N-1}-1\rightarrow N\dagger W^{N-1}+1$ otherwise we would have $N| W^{N-1}\rightarrow (sentence\:7)\: N|W$ false,as we proved Therefore (8) 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 =$$(sentence\:2b) N^2Q+(a-b)^N-AN-a+BN+b\rightarrow N^2| (a-b)^N-AN-a+BN+b$ But: $N^2|X^{N-1}-1\rightarrow N^2|X^N-X=(AN+a)^N-AN-a\rightarrow (sentence\:2b)\:N^2| a^N-AN-a$ and likewise: $N^2|b^N-b-BN$ Therefore: $N^2| (a-b)^N-a^N+b^N$ and accoriding to relations (3) and the uniqueness of (a,b,c) we have proved,we have $b=c\rightarrow N|c-b\rightarrow N|a$ which is false and thus,the asumption: $N\dagger XYZ$ is true no more Last edited by skipjack; January 21st, 2016 at 02:50 PM.
 May 30th, 2013, 11:03 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: A new approach to Fermat's last theorem Most probably false. The thing that identifies the false proofs of FLT are 1. Use of only elementary arithmetic. 2. Too many variables in the proof are being used. 3. Fermat's little theorem used in the proof by contradiction. The proof satisfies all of the three.
 May 30th, 2013, 12:15 PM #3 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 Re: A new approach to Fermat's last theorem More likely you're right, but I would appreciate a more specific reason. Last edited by skipjack; January 21st, 2016 at 02:52 PM.
 May 30th, 2013, 04:45 PM #4 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: A new approach to Fermat's last theorem First, congratulations on not offering up a three line proof based on an immediately obvious misconception. The flipside, however, is that going line by line through a longer proof is both more tedious and difficult when you don't know what the basic thrust is going to be. That's why serious mathematics papers, and academic papers more generally, include at least one abstract at the top outlining the basic logic of what is to be presented. Could you perhaps do that for us?
 May 31st, 2013, 02:03 AM #5 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 Re: A new approach to Fermat's last theorem At first,thanks to Johnr for his kind words.As for his request,this is what I have to say: In an attempt to a simpler approach to the 1st case of Fermat's last theorem $(X^N+Y^N=Z^N (1))$,I split the topic in three steps In the first step,basic divisibility relations between X,Y,Z and N were established.Most of them,if not all,are familiar to those who have pored over the theorem In the second and longest step,I consider the remainders of X,Y,Z divided by N and I examed what their properties must be to satisfy (1).This led to information about the nature of X,Y,Z and N itself. In the third and last step,using the difference X-Y,I showed that N must divide one of X,Y,Z and so the first case of the theorem can't hold I had to keep the level of maths used as elementary as possible ("high school" maths and not "high class" as i mistyped in my topic) Willing to any remarks or errors detecting
 May 31st, 2013, 05:22 AM #6 Senior Member   Joined: Apr 2013 Posts: 425 Thanks: 24 Re: A new approach to Fermat's last theorem Hello! An idea: If $x,y,z\in Z*$ and $n>1$ then necessarily $x, y, z$ are the sides of a triangle.What relationship exists between the sides of a triangle? Thank You!
 May 31st, 2013, 07:24 AM #7 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 Re: A new approach to Fermat's last theorem $|x-y| I wonder what comes next
 May 31st, 2013, 09:19 AM #8 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 Re: A new approach to Fermat's last theorem In my 30-5-13 topic,step 2,we came to the conclusion $N|a^2+ab+b^2$ (1) assuming only that: $N\dagger a$ and of course $X^N+Y^N=Z^N$ If we now assume that N|Y, could (1) tempt us to consider the case ?|????
May 31st, 2013, 09:23 AM   #9
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by bruno59 $|x-y| I wonder what comes next
In the $\Delta ABC$ with sides $z>y>x$ we can write $xcosB+ycosC=z$ , $x^2+y^2-2xycosA=z^2$.It looks easy that $\Delta ABC$ has all sharp angles.
Thank You!

May 31st, 2013, 09:01 PM   #10
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by Dacu Hello! An idea: If $x,y,z\in Z*$ and $n>1$ then necessarily $x, y, z$ are the sides of a triangle.What relationship exists between the sides of a triangle? Thank You!
Wonder how this relates to the OP.

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