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May 30th, 2013, 10:32 AM   #1
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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:

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. .
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 was'nt 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:

From Fermat's little theorem:

and as N doesn't divide R,by sentence 5 it follows:
(1)
On the other hand:

Likewise

Combining these last three equalities we get:
(2)

Step 2 . .
Consider the three Euclidean divisions

with the three (positive) remainders a,b,c being smaller than N.Then:

In any case,it's easy to show that:
(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 (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=6q-1.
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:


and so either:

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,say:

then

which is false
Therfore we assume,say:

Now if N=6q-1,by sentence 1a we would have

which is false,because N doesn't divide either a-b 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.

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 ? does't divide X,Y or Z must be abandoned (quod erat demonstrandum)
Let W=X-Y and so:
(5)
From relation (1):

and because (see relation (2)):

it follows that:


and eleminating denominators:


(6)
???? N can't divide W, or (5) would yield N|3XY and since N doesn't divide 3 we conclude (sentence 5)

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 and it goes like this:
In the first case of Fermat's last theorem,if:

then:

Using this and (7) we immediately get:
(8)
But by Fermat's little theorem:

otherwise we would have

false,as we proved
Therefore (8) gives:

But:

and likewise:

Therefore:

and accoriding to relations (3) and the uniqueness of (a,b,c) we have proved,we have

which is false and thus,the asumption:

is true no more

Last edited by skipjack; January 21st, 2016 at 02:50 PM.
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May 30th, 2013, 11:03 AM   #2
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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.
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May 30th, 2013, 12:15 PM   #3
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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.
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May 30th, 2013, 04:45 PM   #4
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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?
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May 31st, 2013, 02:03 AM   #5
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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 ,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
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May 31st, 2013, 05:22 AM   #6
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Re: A new approach to Fermat's last theorem

Hello!
An idea:
If and then necessarily are the sides of a triangle.What relationship exists between the sides of a triangle?
Thank You!
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May 31st, 2013, 07:24 AM   #7
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Re: A new approach to Fermat's last theorem


I wonder what comes next
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May 31st, 2013, 09:19 AM   #8
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Re: A new approach to Fermat's last theorem

In my 30-5-13 topic,step 2,we came to the conclusion
(1)
assuming only that:

and of course

If we now assume that N|Y, could (1) tempt us to consider the case ?|????
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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

I wonder what comes next
In the with sides we can write , .It looks easy that has all sharp angles.
Thank You!
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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 and then necessarily 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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