The Pellian Equation x2 −Dy2 = 1 from Multiple Sequences
(Part XFa)

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

The Pellian equation is the Diophantine equation x2 − Dy2 = z2 where z equals 1. The least solutions of the Pell equation 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,936,1), where the D values 69 and the 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 somewhat complex sequence but in reality is a mixture of three sequences which as the saying goes is readily amenable to paper and pencil arithmetic with a little work:

0, 0, 69, 38590, 41923, 69300, 73743, 211318, 219023, 219025, 226865, 441485,
452594, 534725, 546944, 860615, 876094, 876098, 891715, 1282428, ...

Since one equation cannot capture all the numbers in the sequence the single sequence can be split into three different paired sequences. The more complex sequence, Pa(n), consists of four F expressions with an initial F while the second, Pb(n), consists of two F expressions with an initial F. The third P(m) is the simplest consisting of a pair of equations where 4682 = 219024. In addition, n and m are initially set at 0:

F4n = 69
Pa(n) =
F4n+1 = F4n + 5503(7+32n)     F4n+2 = F4n+1 + 35153(n+1)
F4n+3 = F4n+2 + 5503(25+32n)    F4n+4 = F4n+3 + 15550(2n+1)
F2n+1 = 41923
Pb(n) =
F2n+2 = F2n+1 + 3911(7+14n)     F2n+3 = F2n+2 + 2309(166)(n+1)
P(m) = (m(219024m − 1)), (m(219024m + 1))

where the red color numbers in the original sequence belong to P(m).

The other properties of these sequences are:

Table F shows the various Ds from the three sequences P(n) and P(m) along with their respective x values except for the initial Da(0)=69, x=7775 omitted from the table since the n numbers start at one. All y values are 936.

Table F
n 12345678 9101112
Da(n)3859073743211318226868441485546944860615 8917151282428145819319479601994610
x183871254177430273445823621919692225868321 8838711059967113027313063691321919
Db(n)41923693004525945347251301313143819825880802779719 4312895455928864757586776905
x1916472464016296956844491067743112249715057911560545 1943839199859323818872436641
D(m)2190232190258760948760981971213197121935043803504388 5475595547560578848587884870
x4380474380498760958760971314143131414517521911752193 2190239219024126282872628289

P(n) uses both R(n)a expressions while P(m) uses only the R(n)b where the n2 are varied according to a repetitive pattern. The R(m), however, is a straightforward calculation:

R(n)a = (n1 + 13√D)2 ∕n2 = x + 936√D
R(n)a = R(n)b = (n1 + 13√D ± 288×n2)2 ∕n2 = x + 936√D
R(m) = (468m1 + √D)2 ∕m1 = x + 936√D

where:

n1 is an integer formed by dividing x by 72
For Pa(n) the n1 starts at 3 and increases by repetitive addition of 68,27,68 and 6 to n1
For Pb(n) the n1 starts at 74 and increases by repetitive addition of 21 and 148 to n1
D are the values from the above sequences starting at 69 for Pa(n) and 41923 for Pb(n).
m1 are consecutive integers: 1, 2, 3, 4, 5, ...
the values of adjacent xs in a pair is (2×4682m1 − 1), (2×4682m1 + 1)
D are the values from the above sequence starting at 219023.

The pairs of Da(n) from Table F and their corresponding Equal Expressions are tabulated in Table I. Db(n) and D(m) pairs are tabulated in Part XFb.

Pell Equal Expressions for (x,936,1) Triples

Table I [Pa(n)]
Pell EquationEqual Expressions
x2 − 69y2 = 1 R69 = (108 + 13√69)2 ∕3 = 7775 + 936√69
x2 − 38590y2 = 1 R38590 = (2554 + 13√38590 + 288×71)2 ∕71 = 183871 + 936√38590
x2 − 73743y2 = 1 R73743 = (3530 + 13√73743 − 288×98)2 ∕98 = 254177 + 936√73743
x2 − 211318y2 = 1 R211318 = (5976 + 13√211318)2 ∕166 = 430273 + 936√211318
x2 − 226868y2 = 1 R226868 = (6192 + 13√226868)2 ∕172 = 445823 + 936√226868
x2 − 441485y2 = 1 R441485 = (8638 + 13√441485 + 288×240)2 ∕240 = 621919 + 936√441485
x2 − 546944y2 = 1 R546944 = (9614 + 13√546944 − 288×267)2 ∕267 = 692225 + 936√546944
x2 − 860615y2 = 1 R860615 = (12060 + 13√860615)2 ∕335 = 868321 + 936√860615
x2 − 891715y2 = 1 R891715 = (12276 + 13√891715)2 ∕341 = 883871 + 936√891715
x2 − 1282428y2 = 1 R1282428 = (14722 + 13√1282428 + 288×409)2 ∕409 = 1059967 + 936√1282428
x2 − 1458193y2 = 1 R1458193 = (15698 + 13√1458193 − 288×436)2 ∕436 = 1130273 + 936√1458193
x2 − 1306369y2 = 1 R1947960 = (18144 + 13√1947960)2 ∕504 = 1306369 + 936√1947960
x2 − 1994610y2 = 1 R1994610 = (18360 + 13√1994610)2 ∕510 = 1321919 + 936√1994610

This concludes Part XFa. Continued in Part XFb. Go back to Part XE.

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