Re: Beal's conjecture
- From: "Roman B. Binder" <rbinder@xxxxxxxxxxxxxxxx>
- Date: Fri, 03 Mar 2006 09:16:47 EST
Hi,
Following Your arguments:
Disproof for a^x + b^y = c^z.....(*)I looked at two sources and the conjecture can beread one of two ways:
c have a common
(1) a^x + b^y = c^z and x,y,z > 2 implies a,b, and
factor d > 1(a,c), and (b,c) are
(2) a^x + b^y = c^z and x,y,z > 2 implies (a,b),
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
with my parameter uses in the fact
equality (ka)^x + (kb)^y = (kc)^z
Once such equality falls so also (*)
regards, roman
Ro-bin
.
- References:
- Re: Beal's conjecture
- From: Chip Eastham
- Re: Beal's conjecture
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