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Re: x^2 - Ay^2 =1
Posted:
May 27, 2007 3:51 PM
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Vincenzo Librandi wrote :
> Philippe wrote:
>>>
>>> I suggested A = n^2 - 3 and A = n^2 + n + 1 and starting from
>>> that given expression, find the solutions of the corresponding
>>> Pell's equation.
>>>
>
> For A=n^2-3
> I found it easy, when n=0 (mod.3).
> ...
that is n = 3m + 3 and :
> that is
> A=9m^2+18m+6 (with m>=0)
> Y=2n/3 and X=[(2n^2/3)-1]
No. That is Y = 2m + 2, X = 6m^2 + 12m + 5
A = polynomial in m => X and Y should be functions of *m*
(were simpler if you choosed just n = 3m)
> A=(6,33,78,141,222,321,...)
> ...
That is you solved the case A = 9n^2+18n+6, (same case as 9n^2 - 3)
not A = n^2 - 3 !
(the 'm' could be named whatever you want : u, t, or ... n)
And all other values from A = n^2 - 3, that is
A = (13,22,46,61,97,118,166,193...) ?
It seems for instance that the n = 3m+1 case splits again into
n = 9m+1, 9m+4 and 9m+7 cases,
and then ... split again into modulo 27 cases ? endless ?
Regards.
--
Philippe C., mail : chephip+news@free.fr
site : http://chephip.free.fr/ (recreational mathematics)
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