silly question x^^a + y^^b = z^^c
master1729
but since i like tetration and FLT and Beal , the analogue idea occured to me :
let x,y,z,a,b,c be integers >= 3
has x^^a + y^^b = z^^c got a solution ?
regards
tommy1729
maverik
Why just don't read about Ferma theorem?
Dan Cass
I think by x^^a he means something different from x^a.
Even if x^a was intended, note the different exponents,
so it's not Fermat's Theorem.
maverik
> Even if x^a was intended, note the different exponents,
> so it's not Fermat's Theorem.
Yes, you are right. My bad.
Pubkeybreaker
YES IT IS.
x^^a = x^a1 for some integer a1.
Why is this difficult?????
Pubkeybreaker
Mea Culpa. I did not realize that he intended a,b,c, to be different.
This is, however the same as the so-called Beal conjecture.
KY
Of course, no one knows what you mean by ^^, fuckwit. This, like everything
else you post, is total boring off-topic mathforum crap. Mathforum - a
festering vat of immeasurable idiocy.
"master1729" <tomm...@gmail.com> wrote in message
news:322836952.5.1257960...@gallium.mathforum.org...
> This might be a stupid question - I made up my own brand-new symbol and I
> don't even know how to to use it! Jesus, what a fuckwit I am.
Jonathan Hoyle
> if the symbol ^^ means addition, this is trivial.
>
> Of course, no one knows what you mean by ^^, fuckwit. This, like everything
> else you post, is total boring off-topic mathforum crap. Mathforum - a
> festering vat of immeasurable idiocy.
There's no need for abusive language. The original poster clearly
meant ^^ to mean the tetration operator. Therefore, it seems (to me
anyway) to be a legitimate question.
Jonathan Hoyle
http://www.jonhoyle.com
master1729
^^ is tetration or power tower notation !!
seems that some are confused.
x^^2 = x^x , x^^^3 = x^(x^x) etc
apart from stupid irrelevant insults and misunderstandings is anyone (now) able to post something sensible ?
Pubkeybreaker
I did. x^^a is x^a1 for SOME integer a1. Thus, x^^a + y^^b =
z^^c is
the same as x^a1 + y^b1 = z^c1 for some integers a1, b1, c1. This
is the so-called
BEAL conjecture.
[note that a1 = x^x^x... (a tower of height x-1)] assuming that
x^^a =
x^(x^x^x^x...) where the height of the entire tower is x.
I.N. Galidakis
> On Nov 12, 7:19 am, master1729 <tommy1...@gmail.com> wrote:
>>> this might be a silly question.
>>
>>> but since i like tetration and FLT and Beal , the
>>> analogue idea occured to me :
>>
>>> let x,y,z,a,b,c be integers >= 3
>>
>>> has x^^a + y^^b = z^^c got a solution ?
>>
>>> regards
>>
>>> tommy1729
>>
>> ^^ is tetration or power tower notation !!
>>
>> seems that some are confused.
>>
>> x^^2 = x^x , x^^^3 = x^(x^x) etc
>>
>> apart from stupid irrelevant insults and misunderstandings is anyone (now)
>> able to post something sensible ?
>
> I did. x^^a is x^a1 for SOME integer a1. Thus, x^^a + y^^b =
> z^^c is
> the same as x^a1 + y^b1 = z^c1 for some integers a1, b1, c1. This
> is the so-called
> BEAL conjecture.
I don't think Tommy's conjecture is completely equivalent to the Beal
conjecture. A solution to Tommy's equation gives a solution for the Beal
conjecture (taking a1 = x^^{a-1}, b1 = y^^{b-1} and c1 = z^^{c-1}), but a
solution to the Beal conjecture doesn't necessarily give a solution to Tommy's
equation.
It is conceivable that one may have:
E a1, b1, c1:
x^{a1} + y^{b1} = z^{c1}
but the a1, b1 and c1 may NOT necessarily be decomposable as towers of x, y and
z.
My guess is that nothing is known about this conjecture and it will probably
stay like this, for a long time.
--
Ioannis
master1729
Thank you.
I im waiting for World Wide Wade , Robert Israel and Dave L Renfro to reply.
These are in my not so humble opinion x) the best posters on sci.math and notable for their non-hostality towards tetration.
Of course you and other members of the tetration forum are also potential intresting posters in this thread.
and of course anyone is welcome to post.
btw , i wonder if Ioannis thinks i should post this elsewhere too , for instance on the tetration forum ?
also i had the impression that this question does have an answer ; provable or at least reducible towards total dependance on Beal or other well known unsolved number theory problems.
for this impression i added " silly question " to the topic name , but perhaps its not so simple afterall.
maybe this problem requires new math tools ? or in other words is undecidable with current number theory methods ?
but what is 'new' of course ... thats a difficult question , since all 'new' needs to reduce towards 'old' in order to prove its validity.
i have , as you might have noticed , claimed a RH proof based on somewhat traditional number theory proof methods , an ' elementary proof ' that is relatively short and requires no complicated concepts , an '18 th century proof' as lwalke calls (or quotes) it.
( ' elementary proof ' is a defined math term )
despite that RH proof and its consequences ( other number theory conjectures are proven when based upon RH or by a similar proof technique ) this FLT-variant ( TLTT tommy's last tetration theorem :p ) seems immume to the methods i used to proof RH.
( and btw i wonder if RH relates to FLT in some strange way , though FLT has been proven , im looking for short proofs nowadays , basicly by motivation of my relatively short RH proof. Probably not. Never read such a thing , no particular argument to think so , counterintuitive , ... but still who knows ? )
regards
tommy1729
master1729
..maybe ...