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February 24th, 2014, 06:43 PM   #41
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Re: Idea of solving FLT n=3

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No one gave enough attention to my post Diophantine equation : you do not know what this post is hiding.
If you post more details, I'd be happy to comment.

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February 24th, 2014, 09:30 PM   #42
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Re: Idea of solving FLT n=3

mobel wrote:No one gave enough attention to my post Diophantine equation : you do not know what this post is hiding.
If you post more details, I'd be happy to comment.

You probably not read the other post too:

If you read well my post you understand that my sum start from the first simple n=2 Diophantine equation:

How to sqaure a square with a triangle: 1,3,5,7...2n-1

Any square is a sum of all the previous odds

And this is well known

What i think is not so known (to amateur) is that this is a general formula where any power of integer is:

A^n = sum of Mn = X^n-(X-1)^n odds numbers where x is the index that start from 1 and arrive to A

So a power of integer is the sum of SPECIAL ODDS (just). All that thanks to Newton's develope.

So as much as the power rise, ...as less as odds number I've to use

What is at all unknown (for me amateur) is that power of integers has another interesting property:

To develope a n-power of integers with integers only (with one rule that works also for all the other powers) you need n indipendent terms.

So there are only this way to develope:

A^n = A*A*A

A^n = (P-1)^n = (Q+1)^n =

where Mn={X^n-(X-1)^n} has exactly n terms

or, IF AND ONLY IF A=k *a if k<1:

Mn/k ={X^n-(X-1)^n}/k if k<1

A^n = (P-k)^n = (Q+k)^n =

or, IF AND ONLY IF A=k *a if k>1:

Mnk ={X^n-(X-1)^n}* k

So, thanks to Newton, we cannot cut in other ways, or it signify that n over k combinations has other possible develope, and is proven that it has not.

So if A^n+ B^n respect Newton's rules it is equal to C^n with A,B,C integers, and is developable in n term.

but if it respect Newton rules it meand that we can work with A,B,C in the same way:







so we can remove the similar terms from:



then:



and finally have



so is this possible and when ?

take look just at right, van be:



A power of integer?

We explain that for Newton's rules a power of integers can be written as a sum

JUST if it is a sum of special odds Mn = {X^n-(X-1)^n} that start from 1 and rise till A (or B or C or else) STEP 1.

OR

JUST if A=k*a (k,a both integers) and if it is a sum of K (or a) special odds Mnk = [{X^n-(X-1)^n}]*k , that start from k and rise till A (or B or C or else) STEP K (or a).

so sum at right :



START FROM B+1, and is a standard step 1 sum so for what above:

- this cannot be a power of integer since a power of integer can be expressed by a sum:

-from 1 to A (or C or else) step 1

- from a value k (hereup Fermat ask k=B+1) to A only if it jumps step K till A = a+k with a,k integers...

But one case rest possible (remember n>2):

if C=B+1 so the step sum 1 or K not matter, for so the only possible case is not a SUM, BUT JUST ONE SPECIAL ODD: {X^n-(X-1)^n} with x=C so to satisfy FERMAT n>2 A,B,C integers, coprime we must have:



we can now have 2 case:

1) {C^n-(C-1)^n = prime than FOR sure it has no solution since {X^n-(X-1)^n} must be divisible by ALL the factor of A

2) C^n-(C-1)^n = non prime so we can check directly if it can be divisible by ALL the factor of A

Remembering we always talking of n>2

for n=3 we can make C as unknown x, so x=C and again:

3x^2-3x+1 - A^3 =? 0 with x integer ?

...of course not since the solution of the 2th degree equation lives a non square under the square root.

We can probably use the analitic solution of n=5 and n=7 equations...

But to get the n>3 general solution we have to look more deep inside infinitesimal calculus....

Or remember the Rational root theorem (from Wiki)

In algebra, the rational root theorem (or rational root test) states a constraint on rational solutions (or roots) of the polynomial equation

a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\!

with integer coefficients.

If a0 and an are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies

p is an integer factor of the constant term a0, and
q is an integer factor of the leading coefficient an.
************************************************** ************************************************** **********

SINCE WE HAVE JUST TO PROVE n= PRIME CASE,

WE KNEW THAT our a0 in Mn|C =(C^n-(C-1)^n is the Newton's develope 2th coefficient = n (or 2th term of tartaglia's Triangle)

We have that a0= n = PRIME =is equal to p in the theorem above... for so n= p = PRIME so X= p/q has no integer solutions FOR ANY P PRIME.... since the trivial p=q is not accepted.

Thanks
ciao
S.

sorry is late... I try to check again...
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February 28th, 2014, 10:56 PM   #43
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Re: Idea of solving FLT n=3

Sorry for the too short conclusion above that is for sure wrong in this terms:

"We have that a0= n = PRIME =is equal to p in the theorem above... for so n= p = PRIME so X= p/q has no integer solutions FOR ANY P PRIME.... since the trivial p=q is not accepted"

What we know is that we have to proove only the case where n= prime so that for:

3 3 1 n=3

5 10 10 5 1 n=5

7 21 35 35 21 7 1 n=7

11 55 165 330 462 462 330 165 55 11 1 n=11

The condition of the rational polinomial solution is respected, and that the solution, if any, must be not just rational, but integer.

So we can note that if n= prime, all the inthernal Tartaglia's coefficient are divisible by n. So what above can be rewritten as:


a0= 3/3 a1= 3/3 a3= (1-A^3)/3 n=3

5/5 10/5 10/5 5/5 (1-A^5)/5 n=5

etc...

so in general: a0= 1, an= (1-A^n)/n

so x=p/q rule becomes:

x= (1-A^n)/n

and x must be an integer.

Now for Fermat the little we know that:

1) A^n/n has the same rest of A/n,

1a) A^n/n has rest 0 if A=k*n.

1b) A^n/n has rest <>0 if A<>k*n.

for so A^n / n in the case

1a) n>2 A>2 ; A<> K*n has for sure a Rest we call Rn >=1

if the rest Rn is >1 than ( A^n -1) / n has a rest Rx = Rn-1 >= 1 so it means that

x= ( A^n -1) / n has a rest Rx, so no integer solution

1b) if the rest Rn is =0 than ( A^n -1) / n has a rest Rx = Rn-1 = -1 that return in 1a)

1c) if the rest Rn is =1 than ( A^n -1) / n has a rest Rx = Rn-1 =0 that is what we have to disproove

In this case we must have

A^n / n with rest Rn=1

( A^n -1) / n with Rx=0

etc....

Thanks
ciao
S.
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