Triangular Numbers and the Pellian Equation x2 −Dy2 = 1 (Part XX)

A New Sequence Sn for the Pellian

The Pellian equation x2 − Dy2 = 1 was covered in Part I and the negative Pellian equation x2 − Dy2 = −1 in Part II. The least solutions of the negative Pell equation, however, are not posted in either Wikipedia (which has a small section describing this topic) or listed in Recreations in the Theory of Numbers by Albert H. Beiler (1966) as were their positive Pell solutions, but the following equations on page 253 may be used for their computation:

x = [(p + qD)2n-1 + (p − qD)2n-1 ∕ 2]
y = [(p + qD)2n-1 + (p − qD)2n-1 ∕ 2D)]

In addition, the method of converting a quadratic surd D into continued fractions (pages 261-262) are also methods that can be used to generate these least solutions. On the other hand, this method is now available on this website as a computer program making it extremely easy to generate the convergents. As to the topic of this article, it concerns the sequence Sn generated from the the equation 36(n)(n+1) − 9 (not in the OEIS database):

Sn = 36(n)(n+1) − 9
-9, 63, 207, 423, 711, 1071, 1503, 2007, 2583, 3231, 3951, 4743, 5607, 6543, 7551, 8631, 9783, 11007...

This sequence is of the type N(n)(n+1) + m employed in Part XIX but differs in that the D values entered into the Pell calculator program produce Qn = +1 at three different n's instead of the same n as was shown in Part XIX. These ns may be considered as a triple (9, 5, 13) where the least solution all fall at the first n − 1 of Qn. Although 9 and 13 give the desired solutions, it is the second least solution of 5, the one where n = 9 that is of interest, since the numbers at n − 1 of Qn match up with those of n = 9 and n = 13. Code 711 shows the calculated convergents for D = 711 with its Qn values of 1 at 5, 9, 13, 17, etc. This may be compared to the other n terms in the triple 9 and 13 Code 423 and Code 1071 where again the values of x and y fall to the left of Qn at n − 1.

Table I shows the first 2-15 D terms in the sequence Sn not including the −9. As mentioned above every three n is a triple (9,[5,9],13) modified to show that it is the 2nd term in the bracket that is to be used. Moreover, the first with a Qn = 3 and the second, the odd man out, are included since both fit the Pellian equation but don't fit the triplet rule.

Table I
D632074237111071150320072583323139514743 560765437551
x811514607127992879956447100351165887259199387199557567 77875110599671411199
y18022448088014562240326445606160809610400 1310416240
n3995,91395,91395,913 95,913

The Pell equations are set up according to the following formats with the x and ys taken from Table I. The first equation with 8(nt)2 and n2 gives the alternate expression for x + yD where nt is a triangular number equal to (n2 + n)/2 and n2 = y/16

RD = (8(nt)2 + n2D)2 ∕ (nt)2 = x + yD

Table II gives the formulas for all 13 Pell equations according to the format listed above where both formulas on the right are equivalent. Entry one with D = 207 is added here to show that n2 = 3 is the second triangular number although the equation is not part of the group as stated previously. The values for pn and qn were obtained from the first least solution at n = 8, n = 8 and n = 12, respectively, for each triple. The triangular numbers are in blue.

Table II
Pell EquationEqual Expressions
x2 − 207y2 = 1 R207 = (32×8 + 5207)232 = 1151 + 80207
x2 − 423y2 = 1 R423 = (62×8 + 14423)262 = 4607 + 224423
x2 − 711y2 = 1 R711 = (102×8 + 30711)2102 = 12799 + 480711
x2 − 1071y2 = 1 R1071 = (152×8 + 551071)2152 = 28799 + 8801071
x2 − 1503y2 = 1 R1503 = (212×8 + 911503)2212 = 56447 + 14561503
x2 − 2007y2 = 1 R2007 = (282×8 + 1402007)2282 = 100351 + 22402007
x2 − 2583y2 = 1 R2583 = (362×8 + 2042583)2362 = 165887 + 32642583
x2 − 3231y2 = 1 R3231 = (452×8 + 2853231)2452 = 259199 + 45603231
x2 − 3951y2 = 1 R3951 = (552×8 + 3853951)2552 = 387199 + 61603951
x2 − 4743y2 = 1 R4743 = (662×8 + 5064743)2662 = 557567 + 80964743
x2 − 5607y2 = 1 R5607 = (782×8 + 6505607)2782 = 778751 + 104005607
x2 − 6543y2 = 1 R6543 = (912×8 + 8196543)2912 = 1059967 + 131046543
x2 − 7551y2 = 1 R7551 = (1052×8 + 10157551)21052 = 1411199 + 162407551

This concludes Part XX.

Go back to homepage.


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