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December 7th, 2015, 01:35 PM   #1
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Wink A conclusive search in Fermat's L.T.

A CONCLUSIVE SEARCH IN FERMAT’S L. T.

Unlike a pointless search for a contra case that has been proved to be non existent, this search which has been with the aim of finding the least magnitude value for

D = A^p - B^p - C^p for all integer values of p above 2, has led to significant conclusions.

The following conditions were applied during the search:

1. A B and C all different positive integers

2. No common factor between any two of A B & C

3. One of A B & C to be even

4. The least value of A to equal (1/ ( p^½-1) +1) rounded up to the nearest integer

5. C = A^p - B^p - C^p - [( A^p - B^p - C^p )^1/p rounded up/down to the nearest integer ]^p

Although the value of D appears to have a random connection with A B and C, it was found that with A fixed and B varied, the magnitude of D is always at a minimum when A - B = 1.

Then with A - B = 1, was found that D is always a minimum when B = 2.p and then C = 2^p - 1, A = 2^p + 1, and the conditions above are always met.

Since all values of p can’t be examined, an algebraic study as follows is needed to show that D can never = zero

Applying binomial expansion to these terms it is seen that with p odd, then D.mod(p) is 2 with all values of p above 2, and then with p odd or even, D has only one negative term (B)p , which is cancelled by a positive term p(B)^(p-1) , leaving p-2 positive terms.

Setting A = 2.p + 1, B = 2.p and C = 2.p - 1 is relevant even to p = 1 or 2 when A is at a minimum.

P D

1 0 } - With p less than 3 there are an infinite number of

2 0 } predictable sets of A, and B that have these values of D

3 2 } These are with A at the minimum.

4 64

5 2002 see note below

6 69264

With p = 5 there is a smaller minimum of 12 with A, B, C = 17,16,13 and this has the surprising result that D also = A - B - C.. Rather like a perfect number in a 5th dimension. Note that D.mod(p) = 2 still applies.

Is it possible that this could occur at some higher value of p? (Conjecture NO)
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December 7th, 2015, 09:38 PM   #2
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I spent lot of time on, but I found numerical check much less interesting than "my" theorem and it's proof:

1.1 - FLT is true also for $\displaystyle A,B,C \in \mathbb Q $

1.2 - The most elegant formulation of FLT trick is:

FLT has no solution for $\displaystyle A,B,C \in \mathbb Q $ for n>=3 since the 1st derivate of y=x^n for n>=3 is a curve.

This becomes clearer if we rewrite the FLT using sums, when we becomes able to manipulate them till the limit, and we note that FLT require this "symmetric" relation for C^n:


$\displaystyle C^n = 2A^n+ \Delta = 2B^n - \Delta$

that is possible just on a linear derivate y=2x.

where for $\displaystyle A,B,C \in \mathbb N $ we can write:

$\displaystyle \Delta = \sum_{X=A+1}^{B} (X^n-(X-1)^n)$

And I've a trick to put this in N than in R.

I've (several) proof for all that, but I need time to let it be presentable since I found all that through my "complicate modulus algebra" that is unfortunately not so easy to be digest from those won't accept the new property I discover and assign to the sum sign, or for those have to asking me a new symbol for what I call step sum....

In any case, I'm working and studying, because I won't be forgotten (I know this from several years)... or remembered as a one of the millions of stupid who lost his life in too big "thinks" he can't understood...

Beal proof follows as minor proof once understood that sum/power's property.

Thanks
Ciao
Stefano

Last edited by skipjack; July 10th, 2017 at 07:38 AM.
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December 12th, 2015, 11:24 AM   #3
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Quote:
Originally Posted by complicatemodulus View Post
I spent lot of time on, but I found numerical check much less interesting than "my" theorem and it's proof:

1.1 - FLT is true also for $\displaystyle A,B,C \in \mathbb Q $

1.2 - The most elegant formulation of FLT trick is:

FLT has no solution for $\displaystyle A,B,C \in \mathbb Q $ for n>=3 since the 1st derivate of y=x^n for n>=3 is a curve.

This becomes clearer if we rewrite the FLT using sums, when we becomes able to manipulate them till the limit, and we note that FLT require this "symmetric" relation for C^n:


$\displaystyle C^n = 2A^n+ \Delta = 2B^n - \Delta$

that is possible just on a linear derivate y=2x.

where for $\displaystyle A,B,C \in \mathbb N $ we can write:

$\displaystyle \Delta = \sum_{X=A+1}^{B} (X^n-(X-1)^n)$

And I've a trick to put this in N than in R.

I've (several) proof for all that, but I need time to let it be presentable since I found all that through my "complicate modulus algebra" that is unfortunately not so easy to be digest from those won't accept the new property I discover and assign to the sum sign, or for those have to asking me a new symbol for what I call step sum....

In any case, I'm working and studying, because I won't be forgotten (I know this from several years)... or remembered as a one of the millions of stupid who lost his life in too big "thinks" he can't understood...

Beal proof follows as minor proof once understood that sum/power's property.

Thanks
Ciao
Stefano
Thanks.
We seem to reach the same conclusion from a different perspective.
After 18 years of pondering the subject, and at age 87, I think it is time to put the subject to bed.

Last edited by skipjack; July 10th, 2017 at 07:39 AM.
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December 14th, 2015, 03:22 AM   #4
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Quote:
Originally Posted by magicterry View Post
Thanks.
We seem to reach the same conclusion from a different perspective.
After 18 years of pondering the subject, and at age 87, I think it is time to put the subject to bed.
Thanks, but I think the subject must return at his right place: below the lame leg of the Math table ;-P

Vice versa from what commonly known this is a very important junction point of Number Theory and (Infimus) Calculus... but involve also topology and complex analysis, just changing the point of view...

I'm working to end the job using Lagrange th. to explain what happen to the limit, once I've made my approximation, we know the derivate etc... but I still need time to avoid stupid bugs... to explain why my intervals that are not equals... etc... But still interesting points on the definition of an asymptote are under inspection. In other words ...lot, lot of work...(for free !)

Last edited by skipjack; July 10th, 2017 at 07:41 AM.
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February 16th, 2016, 01:56 PM   #5
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Further research

My research so far has revealed the conditions for the least possible negative value for "d" to be A = 2p+1 B = 2p C= 2p-1
This is found to be true for all powers p, and of course with p = 1 or 2 where d = 0, this is not the only condition, but that for the smallest values of A, B and C.
Now the surprising minimum value of d = 12 with p = 5 is actually the least positive value of d. So this has led me to see what are the conditions for the least positive value of d with other powers. The result seems to be universally that B = A - 1 and C mostly = C - 2, and d is always larger than the absolute value of the least negative value. Here are my findings for least positive d:-

P A B C d -d
1 3 2 1 0 0
2 5 4 3 0 0
3 11 10 7 12 2
4 28 27 17 306 64
5 17 16 13 12 2002
6 15 14 13 96570 69264
7 20 19 17 24210412 2697354
8 19 18 17 1012154976 117899520

I offer my findings as a matter of interest, not to prove anything.

I spaced this table to make it easy to read, but multiple spaces are now reduced to single spaces. Why?
The -d figures are with A, B and C as defined above.

Last edited by magicterry; February 16th, 2016 at 02:06 PM. Reason: add comment re spacing of result table
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February 16th, 2016, 07:24 PM   #6
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Enclosing the lines in code tags will preserve spacing.
Thanks from topsquark
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July 8th, 2017, 05:55 AM   #7
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Final? conclusions

SOME CONCLUSIONS IN THE SEARCH
In the search to find the smallest value for A^p - B^p - C^p for any value p with A,B,C being positive integers relatively prime, the following conclusions have been reached :-

POWER p = 1
The smallest value is zero and there are an infinite number of sets of A,B,C, the smallest of which is 3,2,1. The number of sets possible with a given value for A is int(A/2)

POWER p = 2
The smallest value is zero and there are an infinite number of sets of A,B,C, the smallest of which is 5,4,3. The number of sets (Pythagoras triples) depend on the number of prime factors in A that are of form 4n+1.

POWER p = 3
The smallest value is 2 and there are an infinite number of sets of A,B,C, the smallest of which is 7,5,6. A = 6n^3+1, B = A-2, C =(A^3-B^3-2)^(1/3)

POWER 4
The smallest value is 64 and there are only three sets of A,B,C which are 3,2,1 : 9,8,7 and 37,36,21.

POWER 5
The smallest value is 12 and there is only one set A,B,C = 17,16,13.

POWER> 5
There is only one set of A,B,C = 3,2,1 giving the following minimum values up to power 10:-
power value
6 664
7 2058
8 6304
9 19170
10 58024

For all powers greater than 2 the above conclusions can only be considered as highly probable. All we know for certain is that they can't be zero.
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July 10th, 2017, 06:19 AM   #8
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You will probably save time using the Complicate Modulus method and the revised FLT equation:

$C^n= 2A^n+\Delta$

and:

$C^n=2B^n-\Delta$

with $\Delta=\sum_{x=A+1}^{B}(x^n-(x-1)^n)$

Tabulating (IR= Integer Root of P/2):

$P$ , $IR_1= \lfloor{(P/2)^{(1/3)}}\rfloor$, $Rest=P-IR_1$

and

$P$ , $IR_2= \lfloor{(P/2)^{(1/3)}}\rfloor +1$, $Rest=IR_2-P$

and $(P)^{(1/3)}$ as check

you'll see:

$341_ 5_ 91__ 341_ 6_ -91_ 6,986368028$
$342_ 5_ 92__ 342_ 6_ -90_ 6,993190657$
$343_ 5_ 93__ 343_ 6_ -89_ 7 $

Since (make the complete table to see it) the Maximum Rest for $2A^3$ is littlest than the one for $2B^3$ and the closest one is the one for 5,6 that is +121 Vs -122 and the Integer solution will comes just in case $RestMax=2*\Delta$ so in this case $91*2=182$

This solution allow you to prove that this is a minimum, and
also rising $n$ the difference can just rise....
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