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Re: x^2 - Ay^2 =1
Posted:
Jun 23, 2007 5:41 AM
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continued...
Hi,
we will give to all the polinomials formulas that
give the primitive of Pell's equation x^2-Ay^2=1
**************************************
(25)
If A=(n^2-4)/39^2,y=39n/2,x=(n^2/2)-1
(with n>=340 and (n^2-4)=0 mod.39^2)
n=340
A=76
Y=6630
X=57799
======================================
(26)
If A=(n^2-1)/6^2, y=6,x=n
(with n>=19 and (n^2-1)=0 mod.6^2)
n=(19,35,37,53,55,71,73,...,)
A=(10,34,38,78,84,140,148,...,)
Y=(6,6,6,6,6,6,6,...,)
X=(19,35,37,53,55,71,73,...,)
=====================================
(27)
If A=(n^2+1)/41^2, y=82n, x=2n^2+1
(with n>=378 and (n^2+1)=0 mod.41^2)
n=(378,1303,2059,2984,...,)
A=(85,1010,2522,5297,...,)
Y=(30996,106846,168838,244688,...,)
X=(285769,3395619,8478963,17808513,...,)
========================================
(28)
If A=(n^2+2)/11^2, y=11n, x=n^2+1
(with n>=102 and (n^2+2)=0 mod. 11^2)
n=(102,140,223,261,344,...,)
A=(86,162,411,563,978,...,)
Y=(1122,1540,2453,2871,3784,...,)
X=(10405,19601,49730,68122,118337,...,)
=========================================
(29)
If A=(n^2-1)/21^2, y=21, x=n
(with n>=197 and (n^2-1)=0 mod.21^2)
n=(197,244,440,442,638,685,...,)
A=(88,135,439,443,923,1064,...,)
Y=(21,21,21,21,21,21,...,)
X=(197,244,440,442,638,685,...,)
===================================
(30)
If A=(n^2+1)/53^2, y=106n, x=2n^2+1
(with n>=500 and (n^2+1)=0 mod.53^2)
n=(500,2309,3309,5118,...,)
A=(89,1898,3898,9325,...,)
Y=(53000,244754,350754,542508,...,)
X=(5000001,10662963,21898963,52387849,...,)
===========================================
continued ...
Regards
Vincenzo Librandi
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