The Pellian Equaltion x2 −Dy2 = 1 from two Paired Sequences P(n)/P(m) (Part XD)

A Method for Generating Pellian Triples (x,y,1)

The Pellian Equaltion is the Diophantine Equaltion x2 − Dy2 = z2 where z Equalls 1. The least solutions of the Pell Equaltion are posted in Wikipedia, in Google books:Canon Pellianus with about 1000 entries and on Table 91, page 254 of Recreations in the Theory of Numbers by Albert H. Beiler (1966), where the values for D on page 252-253 have been computed using the following two expressions:

x = [(p + q√D)n + (p − q√D)n ∕ 2]
y = [(p + q√D)n + (p − q√2D)n ∕ 2√2D)]

The tables in the Canon Pellianus article shows a list of numbers corresponding to triples of the type (x,20,1), where seven D values (6-903) and their corresponding x and y values. It will be shown here, as was shown in Part XA for y = 12, that we can continue generating all those values of x not listed in these articles by employing what appears to be a new sequence but in reality is a mixture of two sequences:

0, 0, 6, 57, 99, 155, 308, 398, 402, 501, 759, 897, 903, 1053, 1410, 1536, 1604, 1802, 2261, 2495, 2505, 2751, 3512, 3594, 3606, ...

Since one equation cannot capture all the numbers in the sequence the single sequence can be split into two different paired sequences composed of the following two expressions:

F2 = 6
P(n) =
F2n+1 = F2n + 51(2n-1)   F2n+2 = F2n+1 + 98n
P(m) = (m(100m − 1)), (m(100m + 1))

where the first expression P(n) is composed of a pair of numbers, each number is the sum of the preceding one starting out with an initial value, F2 = 6 and the counter n set to 1. Thus, according to the second line F3 = 57 and F4 = 155 with F4 subsequently used in the next line when n is incremented to 2. The initial pair is consequently (F2,F3) followed by (F4,F5) followed by (F6,F7), etc. consistent with the sequence P(n). As for the second expression, P(m) is treated as a paired sequence which uses a pair of equations to generate the two paired values.

The other properties of these sequences are:

Table D shows the various Ds from the two split sequences P(n) and P(m) along with their respective x values. All y values are 20.

Table D
n 12345678 91011121314
D(n)65715530850175910531410 180222612751331239004563
x49151249351449551649751 8499511049115112491351
D(m)9910139840289790315961604 249525053594360648934907
x199201399401599601799801 99910011199120113991401

Both P(n) and P(m) use the same method but the mathematical expressions are different and involves multiplying the initial least solutions by either of the two parts of the following mathematical expression:

R(n)D = (5n1 + 2√D)2 ∕ ⅕n1 = x + 15√D
R(m)D = (10m1 + √D)2 ∕ 1⁄10m1 = x + 15√D

where:

n1 are the odd integers: 1, 3, 5, 7, 9...
the values of adjacent xs in a pair is (100(2n − 1) − 51), (100(2n − 1) + 51)
D are the values from the above sequence starting at 6.
m1 are consecutive integers: 1, 2, 3, 4, 5, ...
the values of adjacent xs in a pair is (200m1 − 1), (200m1 + 1)
D are the values from the above sequence starting at 99.

The first three pairs of D(n) and D(m) from Table D and their corresponding Equal Expressions are tabulated in Table I.

Pell Equal Expressions for (x,20,1) Triples

Table I
Pell EqualtionEqual Expressions
x2 − 6y2 = 1 R6 = (5 + 2√6)2= 49 + 20√6
x2 − 57y2 = 1 R135 = (15 + 2√135)2 ∕3 = 151 + 20√135
x2 − 99y2 = 1 R439 = (10 + √439)2 = 199 + 20√439
x2 − 101y2 = 1 R443 = (10 + √443)2 = 201 + 20√443
x2 − 155y2 = 1 R923 = (25 + 2√155)2 ∕5 = 249 + 20√155
x2 − 308y2 = 1 R308 = (35 + 2√308)2 ∕7 = 351 + 20√308
x2 − 398y2 = 1 R398 = (20 + √398)2 ∕2 = 399 + 20√398
x2 − 402y2 = 1 R402 = (20 + √402)2 ∕2 = 401 + 20√402
x2 − 501y2 = 1 R501 = (45 + 2√501)2 ∕9 = 449 + 20√501
x2 − 759y2 = 1 R759 = (55 + 2√759)2 ∕11 = 551 + 20√759
x2 − 897y2 = 1 R897 = (30 + √897)2 ∕3 = 599 + 20√897
x2 − 903y2 = 1 R903 = (30 + √903)2 ∕3 = 601 + 20√903

This concludes Part XD. Go to Part XE. Go back to Part XC.

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Copyright © 2021 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com