Re: Beal's conjecture


stush@xxxxxxxxxxxxxx wrote:
Roman B. Binder wrote:
Also, it does not deserve the name 'Beal
Conjecture'. The extension
from the
Fermat conjecture to one of unequal exponents is one
that is obvious to
any number
theorist. And Mr. Beal isn't even a mathematician.

At a number theory conference in Arcata California in
1985, I was
present when John Tate
gave a terrific presentation of the (then brand new)
connection between
Taniyama-Shimura
and FLT that was given by Frey's work. Tate
mentioned that there were
a couple of obstacles,
including a conjecture of Serre, (proved by Ken
Ribet) that stood in
the way. At the end
of the talk a member of the audience asked the
question: Does the
result also apply to the
case of unequal exponents?

Posing the question in such a casual manner shows
that the conjecture
is rather obvious.

If anything, whoever posed the question is more
deserving to have the
conjecture named
after him. I wish I could remember who it was.

Hi,
Fermat used to express some kind of marvelous
demonstration to the case of powers bigger than 2.
Is it written in his statement, that such powers
should be equal ?
There is really something unexpected and common
for equal and not equal powers bigger than 2.
I used to come back to some trivial transformation
and just noticed once overlooked link to proof.
It looks there for very significant simplifications
of this conjecture and why: plotted parameter shows
to be just rational number...

Ro-bin

P.S. Please do not mix me so much with my previous
errors: my properties to square determinants
falls to factorization of X=mk but k=1...
if anybody used to read my previous posts...

I looked at two sources and the conjecture can be read one of two ways:

(1) a^x + b^y = c^z and x,y,z > 2 implies a,b, and c have a common
factor d > 1

(2) a^x + b^y = c^z and x,y,z > 2 implies (a,b), (a,c), and (b,c) are
not all 1.

Which is correct?

I believe these are equivalent. Suppose a,b have a common prime
factor. Then this prime divides c^z and hence must divide c also.

Similarly for any common factor of a,c or b,c...

regards, chip

.

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