Re: --- --- --- Two similar equations
- From: david <dlee753@xxxxxxxxx>
- Date: Wed, 12 Nov 2008 18:29:28 -0800 (PST)
On Nov 12, 12:19 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
david <dlee...@xxxxxxxxx> writes:
On Nov 12, 12:01=A0am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article
<8429f630-0a6e-4dfc-ba49-a032d4e71...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
=A0Pumpledumplek...@xxxxxxxxx wrote:
On Nov 11, 5:20=A0pm, david <dlee...@xxxxxxxxx> wrote:
The following two equations (1) and (2) are similar and satisfy the
given conditions.
x^p + y^p =3D z^2 =A0 =A0 =A0 =A0 =A0 =A0 =A0(1)
x^p + (1/4m)y^p =3D z^2 =A0 =A0 (2)
Conditions: (x, y, z) =3D 1, y is even, p is a prime > 5, m is an odd
integer > 0.
It is known[1] (1) has no integer solutions.
Question: Based on the properties of (1) is it possible to conclude
that (2) also cannot be satisfied if all of x, y, z in (2) are
integers each > 1.
Any helpful comments will be appreciated.
[1] Darmon and Merel, J.Reine Angew,Math.490(1997),81-100.
I'm not sure that this is even true. The result (or some variant of
it) does not follow from
the statement about equation (1), but one could modify the methods of
Darmon and Merel to prove
something about equation (2) with, for instance, m prime and p larger
than a function of m, at least for
"most" primes m....
de Pumpster
Let p =3D 7, x =3D 9, z =3D 32 t + 5 for any t > 70 or so.
Then z^2 - x^p =3D 32 r for some odd r.
Let m =3D r^6, y =3D 2 r.
--t -
Gerry Myerson (ge...@xxxxxxxxxxxxxxx) (i -> u for email)- Hide quoted
tex=
- Show quoted text -
Kindly give a numerical example to show that (2) can have integer
solutions. Otherwise, I will assume like(1) (2) also cannot have any
integer solutions for the given conditions.
With t=70 in Gerry's example:
p = 7, x = 9, z = 2245, r = 8033, m = 268699341295243950331969,
y = 16066.
--
Robert Israel isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada- Hide quoted text -
- Show quoted text -
Your example is convincing. Thank you very much. Now kindly consider
the two situations below
Situation-1: y = (z^2- x^p)^(1/p): Situation-2: y = [4m(z^2 -x^p)]^
(1/p)
Situation-1 tells y cannot be an integer. Now if (m, (z^2-x^p)) = 1
then in Situation-2 y can be an integer.
You have given an excellent numerical example. Kindly offer some
theoretical explanation.
A simple reply will be appreciated
.
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- Re: --- --- --- Two similar equations
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- Re: --- --- --- Two similar equations
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- Re: --- --- --- Two similar equations
- From: Robert Israel
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