The Pellian Equation x2 −Dy2 = ±1 from a Sequence Sn (Part IIIA)
A Method for Generating Pellian Triples (x,y,±1) and (x,y,−1)
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 + q√D)2n-1 + (p − q√D)2n-1 ∕ 2]
y = [(p + q√D)2n-1 + (p − q√D)2n-1 ∕ 2√D)]
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, it has been found that the group of Ds 13, 29 and 53 described in the articles above are actually part of the sequence with the OEIS number A078370. In addition, the numbers in this sequence are a subset of the numbers for which the negative Pell equation is soluble as listed in A031396:
Sn = 4(n)(n+1) + 5
13, 29, 53, 85, 125, 173, 229, 293, 365, 445, 533, 629, ...
where every number in this sequence has been found to be a surd for the regular Pell equation.
The method described here is different from that of Part II and requires a different set of mathematical expressions, viz., R1 and R2 to generate the subsequent x and y values from the numbers in the sequence Sn.
R1 = [(n1 + √D)2 ∕4]3∕2 = (n2 + n3√D)1∕2 = n4 + n5√D
R2 = [(n1 + √D)2 ∕4]3 = (n2 + n3√D) = (n4 + n5√D)2
where
R1 corresponds to ± staggered Pell triples
R2 corresponds to the separate regular or negative Pell triples
n1 equals 2n + 3 corresponding to the values: 3, 5, 7, 9, 11, ...
n2, n3, n4 and n5 are available or calculated values
the last equality in both expressions is the negative Pell
where available means from the two articles above or if not available can be calculated from the first equality of the RD equations.
The Actual Method
R1 is initially multiplied and rounded off with the known least value solutions x and y of the positive Pell equation to generate a new x and y solutions with z = −1. This new row of values is then tested to make sure it fits the negative Pell equation. Further multiplication and rounding off of this row by R1 produces staggered +1 and −1 Pell values. Once the desired number of rows is obtained we can backtrack and generate the approximate least solution values of the negative Pell equation by taking the initial +1 least solutions x and y and dividing by R1.
In addition, to obtain the separate non staggered triples one uses R2, the square of R1.
The first three tables below are tabulated with the actual literature x and y values while in the latter two, the x and y values were calculated from the left hand side of the RD. As an aside these numbers would probably be difficult to come by via any other method.
Note an online search shows that the equation R2 = [(3 + √13)2 ∕4]3 = 649 + 180√13 is listed on page 187 in the pdf article Solving the Pell Equation by H.W. Lenstra Jr.
Tables of D and Pell (x,y,±1) and (x,y,−1) Triples
- Table I shows first triples for the Pell equation x2 − 13y2 = ±1 using the first least solution (649,180,1) and the multiplicand R1 = [(3 + √13)2) ∕4]3∕2 = (649 + 180√13)1∕2 = 18 + 5√13 = 36.02775638.
- Table II shows the triples for the Pell equation x2 − 13y2 = −1 using the first least solution (18,5,-1) and the multiplicand R2 = [(3 + √13)2 ∕4]3 = 649 + 180√13 = (18 + 5√13)2 = 1297.9992295835.
Table I D=13
| x |
y | z |
| 18 | 5 | -1 |
| 649 | 180 | 1 |
| 23382 | 6485 | -1 |
| 842401 | 233640 | 1 |
| 30349818 | 8417525 | -1 |
| 1093435849 | 303264540 | 1 |
| 39394040382 | 10925940965 | -1 |
| 1419278889601 | 393637139280 | 1 |
|
|
Table II D=13
| x |
y | z |
| 18 | 5 | -1 |
| 23382 | 6485 | -1 |
| 30349818 | 8417525 | -1 |
| 39394040382 | 10925940965 | -1 |
| 51133434066018 | 14181862955045 | -1 |
|
- Table III shows first triples for the Pell equation x2 − 29y2 = ±1 using the first least solution (9801,1820,1) and the multiplicand R1 = [(5 + √29)2) ∕4]3∕2 = (9801 + 1820√29)1∕2 = 70 + 13√29 = 140.0071424927.
- Table IV shows the triples for the Pell equation x2 − 29y2 = −1 using the first least solution (70,13,-1) and the multiplicand R2 = [(5 + √29)2 ∕4]3 = 9801 + 1820√29 = (70 + 13√29)2 = 19601.99994894.
Table III D=29
| x |
y | z |
| 70 | 13 | -1 |
| 9801 | 1820 | 1 |
| 1372210 | 254813 | -1 |
| 192119201 | 35675640 | 1 |
| 26898060350 | 4994844413 | -1 |
| 3765920568201 | 699313893460 | 1 |
|
|
Table IV D=29
| x |
y | z |
| 70 | 13 | -1 |
| 1372210 | 254813 | -1 |
| 26898060350 | 4994484413 | -1 |
|
- Table V shows first triples for the Pell equation x2 − 53y2 = ±1 using the first least solution (66249,9100,1) and the multiplicand R1 = [(7 + √53)2) ∕4]3∕2 = (66249 + 9100√53)1∕2 = 182 + 25√53 = 364.00274723.
- Table VI shows the triples for the Pell equation x2 − 53y2 = −1 using the first least solution (182,25,-1) and the multiplicand R2 = [(7 + √53)2 ∕4]3 = 66249 + 9100√53> = (182 + 25√53)2 = 132497.99999098.
Table V D=53
| x |
y | z |
| 182 | 25 | -1 |
| 66249 | 9100 | 1 |
| 24114818 | 3312425 | -1 |
| 8777860001 | 1205731800 | 1 |
| 3195165155182 | 438889687625 | -1 |
|
|
Table VI D=53
| x |
y | z |
| 182 | 25 | -1 |
| 24114819 | 3312425 | -1 |
| 3195165155182 | 438889687625 | -1 |
|
- Table VII shows first triples for the Pell equation x2 − 85y2 = ±1 for which the least solution (285769,30996,1) was calculated and the multiplicand R1 = [(9 + √85)2) ∕4]3∕2 = (285769 + 30996√85)1∕2 = 378 + 41√85 = 756.0013228.
- Table VIII shows the triples for the Pell equation x2 − 85y2 = −1 using the first least solution (378,41,-1) and the multiplicand R2 = [(9 + √85)2 ∕4]3 = 285769 + 30996√85 = (378 + 41√85)2 = 571537.99999825.
Table V D=85
| x |
y | z |
| 378 | 41 | -1 |
| 285769 | 30996 | 1 |
| 216041742 | 23433017 | -1 |
| 163327842721 | 17715391848 | 1 |
|
|
Table VI D=85
| x |
y | z |
| 378 | 41 | -1 |
| 216041742 | 23433017 | -1 |
|
- Table IX shows first triples for the Pell equation x2 − 125y2 = ±1 for which the first least solution (930249,83204,1) was calculated and the multiplicand R1 = [(11 + √125)2) ∕4]3∕2 = (930249 + 83204√125)1∕2 = 682 + 61√125 = 1364.00733.
- Table X shows the triples for the Pell equation x2 − 125y2 = −1 using the first least solution (682,61,-1) and the multiplicand R2 = [(11 + √125)2 ∕4]3 = 930249 + 83204√125 = (682 + 61√125)2 = 1860497.99999463.
Table IX D=125
| x |
y | z |
| 682 | 61 | -1 |
| 930249 | 83204 | 1 |
| 1268860318 | 113490317 | -1 |
| 1730726404001 | 154800875592 | 1 |
|
|
Table X D=125
| x |
y | z |
| 682 | 61 | -1 |
| 1268860318 | 113490317 | -1 |
|
This concludes Part IIIA. To continue to Part IIIB.
Go back to Part II.
Go back to homepage.
Copyright © 2020 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com