Re: Full Beal Conjecture (revised)
carlyboxx_at_yahoo.com
Date: 12/15/04
- Next message: Gregory L. Hansen: "Re: Science, Philosophy, Mysticism, Art, Mathematics, and Physics"
- Previous message: LD: "Re: Ring Isomorphism"
- Messages sorted by: [ date ] [ thread ]
Date: 15 Dec 2004 09:30:08 -0800
I was doing a little research on the FLT and Beal conjecture thanks for
the insight on Beal conjecture and Fermat's Last.
> Hi,
>According to great interest at this question,
>I would like to discuss following classification:
>Once generally Beal Conjecture dissmiss x^a + y^b = z^c
>for x;y;z integers of gcd=1 and for a;b;c natural numbers >2
>1.) For a;b;c as odd numbers we'll have some fall of Beal's
>conjecture, which might be proved once using similar method
>as aplicable for FLT conjecture for p=3 and bigger primes.
> ( this method uses also factorisation of numbers shaped as 3M+2
> into factors as square numbers, what is applicated as 6n+1
> shapes and turns into squares of all primes once beginning
> with 5 . I used to write down first lines of it at ca 17.02.04 )
>2.) For one, two or even three of exponents a;b;c as even numbers
>but bigger than 4 and so on containing some odd number bigger or equal
to 3 it will be possible >to turn such fall into (1)-st one: a=a1*a2;
b=b1*b2; c=c1*c2 so (x^a2)^a1 + (y^b2)^b1 = > > > > > (z^c2)^c1 where
a1;b1;c1 odd numbers.
> 3.) For a;b;c as even numbers but discribed exclusively as 2^k or
with the part a2;b2;c2 as 2^k >where k=2 or bigger natural numbers so
such fall is to collect into x^4 + y^4 = z^4 conjecture
> 4.) For only few combinations once one of a;b;c is equal to 4 but
others odd or two of a;b;c >are equal to 4 but the one resisting odd
there is not possible to use the classification and >method for odd
exponents.
>5.) Finally for a;b;c as even numbers but shaped in that way,
> that it is possible to rewrite conjectures:
> x^2 + y^4 = z^4 or x^4 + y^4 = z^2
> You can see, that conjectures x^4 + y^4 = z^4
>and x^2 + y^4 = z^4 or x^4 + y^4 = z^2 were already determinated
>firstly by P.Fermat. Once it is some hope for method for odd exponents
so the resisting small >gap can be closed too and Beal conjecture
dissmiss x^a + y^b = z^c
>> for x;y;z integers of gcd=1 and for a;b;c natural numbers >2
>it will be possible to express that any sum of powers bigger
>than 2 can not be substituted with the power bigger than 2.
> (Does Fermat expressed, that exponents of such powers are equal ?)
> Enjoy super games of powers
>> Hi,
>> According to great interest at this question,
>> I would like to discuss following classification:
>> Once generally Beal Conjecture dissmiss x^a + y^b = z^c
>> for x;y;z integers of gcd=1 and for a;b;c natural numbers >2
>> 1.) For a;b;c as odd numbers we'll have some fall of Beal's
>> conjecture, which might be proved once using similar method
>> as aplicable for FLT conjecture for p=3 and bigger primes.
>> ( this method uses also factorisation of numbers shaped as 3M+2
>> into factors as square numbers, what is applicated as 6n+1
>> shapes and turns into squares of all primes once beginning
>> with 5 . I used to write down first lines of it at ca 17.02.04 )
>> 2.) For one, two or even three of exponents a;b;c as even numbers
>> but bigger than 4 and so on containing some odd number bigger or
equal to 3 it will be >possible to turn such fall into (1)-st one:
a=a1*a2; b=b1*b2; c=c1*c2 so (x^a2)^a1 + (y^b2)^b1 = > (z^c2)^c1 where
a1;b1;c1 odd numbers.
> 3.) For a;b;c as even numbers but discribed exclusively as 2^k or
with the part a2;b2;c2 as >2^k where k=2 or bigger natural numbers so
such fall is to collect into x^4 + y^4 = z^4 >conjecture
> 4.) For only few combinations once one of a;b;c is equal to 4 but
others odd or two of a;b;c >are equal to 4 but the one resisting odd
there is not possible to use the classification and >method for odd
exponents.
> 5.) Finally for a;b;c as even numbers but shaped in that way,
> that it is possible to rewrite conjectures:
> x^2 + y^4 = z^4 or x^4 + y^4 = z^2
> You can see, that conjectures x^4 + y^4 = z^4
> and x^2 + y^4 = z^4 or x^4 + y^4 = z^2 were already determinated
> firstly by P.Fermat. Once it is some hope for method for odd
exponents so the resisting small >gap can be closed too and so
> on it will be possible to express that any sum of powers bigger
> than 2 can not be substituted with the power bigger than 2.
> (Does Fermat expressed, that exponents of such powers are equal ?)
>I appreciate your intent to help but I can't understand your (math)
>grammer. Sorry and thanks again. Kerry
>> Hi,
> I'll cut the grammer and hope You'll anderstand:
>> Beal conjecture dissmiss x^a + y^b = z^c
>> for x;y;z integers of gcd=1 and for a;b;c natural numbers >2
> 1
>.Fall): For a;b;c as odd numbers can be proved once using such method,
wchich is also correct
>for FLT conjecture for p=3 and bigger primes.
>> ( this method uses also factorisation of numbers shaped as 3M+2
>> into factors as c^2 (squares), what is applicated to 6n+1
>> shapes and turns into squares of all primes once beginning
>> with 5 . I used to write down first lines of it at ca 17.02.04 ;
see topic "Square factors >of the odd values of shape 3M+2")
> 2.Fall) For one, two or
>three of exponents a;b;c as even numbers
>but bigger than 4 and so on containing some odd number bigger or equal
to 3 it will be possible >to turn such fall into (1
>Fall)one: a=a1*a2; b=b1*b2; c=c1*c2 so (x^a2)^a1 + (y^b2)^b1 =
(z^c2)^c1 where a1;b1;c1 odd >numbers.
> 3
>.Fall)
>For a;b;c as even numbers but discribed exclusively as 2^k or with the
part a2;b2;c2 as 2^k >where k=2 or bigger natural numbers
>
>it will turn into x^4 + y^4 = z^4 conjecture, solved firstly
>by P. Fermat
>
> 4.Fall)
>For only few combinations once one of a;b;c is equal to 4 but others
odd or two of a;b;c are >equal to 4 but the one resisting odd there is
not possible to use the
>
>method for odd exponents.
>It turns into conjectures: x^4 + y^b = z^c where b;c equal 3
> or bigger prime numbers
> x^4 + y^4 = z^c where c equal 3 or
> bigger prime numbers
>with possible inversions of x;y;z values
>Once generally Beal Conjecture dissmiss x^a + y^b = z^c
>for x;y;z integers of gcd=1 and for a;b;c natural numbers >2
> 5.Fall)
>Finally for a;b;c as even numbers but shaped in that way,
> that it is possible to rewrite
>it into
>conjectures:
>> x^2 + y^4 = z^4 or x^4 + y^4 = z^2
> what used to be profed also
>by P. Fermat
>Once it is some hope for method for odd exponents so the resisting
small gap can be closed too >and so on it will be possible to express
that any sum of powers bigger than 2 can not be >substituted with the
power bigger than 2.
> (Does Fermat>
>express,
>that exponents of such powers are equal ?)
>>I appreciate your intent to help but I can't understand your (math)
>>grammer. Sorry and thanks again. Kerry
> So do I for Your interest and apologise for my first edition
>grammar poverties. I hope, that You can understand my position
>litlle bit better...
- Next message: Gregory L. Hansen: "Re: Science, Philosophy, Mysticism, Art, Mathematics, and Physics"
- Previous message: LD: "Re: Ring Isomorphism"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
