The Pellian Equation x² −Dy² = ±1 Continuation (Part IIIB)

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.

Furthermore, though R₃ and R₄ with D values ranging from 41 through 113 may be expressed via two expressions unlike R₁ and R₂ in Part IIIA which exist also in a cubed form. Thus, the reason for treating these expressions separate from Part IIIA. In addition, while this part and Part II can generate negative Pellian triples they differ in that y is constant for the negative Pell in Part II but variable here. The expressions for generating the triples are:

R₃ = a + b√D = (n₃ + n₄√D)^1∕2
R₄ = (a + b√D)² = n₃ + n₄√D

where a + b√D is in actuality the negative Pell equation which we can generate from the regular Pell equation via exhaustive subtraction of D×a square from n₃ until n₃ is a perfect square. The negative Pell can also be constructed during the construction of Pell triples as described below.

In addition, a method of generating a sequence of Ds which can be used in these Pell equations has not been found to date and the OEIS database did not provide a definitive answer.

The Actual Method

R₃ or R₄ 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 either of these Rs 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 the R being used.

It is simple and generates the least solution of the negative Pell equation which we can tabulate in a new table by either extracting the x and y values from the staggered table or by creating the new table from scratch.

Tables of D and Pell (x,y,±1) and (x,y,−1) Triples

Table VI shows first triples for the Pell equation x² − 41y² = ±1 using the first least solution (2049,320,1) and the multiplicand R₃ = 32 + 5√41 = (2049 + 320√41)^1∕2 = 64.015621187164.
Table VII shows the triples for the Pell equation x² − 41y² = −1 using the first least solution (32,5,-1) and the multiplicand R₄ = (32 + 5√41)² = 2049 + 320√41 = 4097.9997559785. The same multiplicand is used to generate the Pell equation x² − 41y² = +1. Table not shown.

Table VI D=41
x	y	z
32	5	-1
2049	320	1
131168	204855	-1
8396801	1311360	1
537526432	83947525	-1

Table VII D=41
x	y	z
32	5	-1
131168	204855	-1
537526432	83947525	-1

Since the mixed tables contain both positive and negative Pell values the negative Pell will no longer be tabulated according to the example above.
Table VIII shows first triples for the Pell equation x² − 58y² = ±1 using the first least solution (19603,2574,1) and the multiplicand R₃ = 99 + 13√58 = (19603 + 2574√58)^1∕2 = 198.0050504.
Table IX shows first triples for the Pell equation x² − 61y² = ±1 using the first least solution (1766319049,226153980,1) and the multiplicand R₃ = 29718 + 3805√61 = (1766319049 + 226153980√61)^1∕2 = 59436.0000168.

Table VIII D=58
x	y	z
99	13	-1
19603	2574	1
3881493	509665	-1
768555217	100916244	1
152177814459	19981925977	-1

Table IX D=61
x	y	z
29718	3805	-1
1766319049	226153980	1
104982939026082	13441687969085	-1

Table X shows first triples for the Pell equation x² − 73y² = ±1 using the first least solution (2281249,267000,1) and the multiplicand R₃ = 1068 + 125√73 = (2281249 + 267000√73)^1∕2 = 2136.000468.
Table XI shows first triples for the Pell equation x² − 74y² = ±1 using the first least solution (3699,430,1) and the multiplicand R₃ = 43 + 5√74 = (3499 + 430√74)^1∕2 = 86.01162634.

Table X D=73
x	y	z
1068	125	-1
2281249	267000	1
4872748932	570312125	-1

Table XI D=74
x	y	z
43	5	-1
3699	40	1
318157	36985	-1

Table XII shows first triples for the Pell equation x² − 89y² = ±1 using the first least solution (500001,53000,1) and the multiplicand R₃ = 500 + 53√89 = (500001 + 53000√89)^1∕2 = 1000.0009999.
Table XIII shows first triples for the Pell equation x² − 97y² = ±1 using the first least solution (62809633,6377352,1) and the multiplicand R₃ = 5604 + 569√97 = (62809633 + 6377352√97)^1∕2 = 11208.000089.

Table XII D=89
x	y	z
500	53	-1
500001	53000	1
500001500	53000053	-1

Table XIII D=97
x	y	z
5604	569	-1
62809633	6377352	1
703970372268	71477361785	-1

Table XIV shows first triples for the Pell equation x² − 106y² = ±1 using the first least solution (32080051,3115890,1) and the multiplicand R₃ = 4005 + 389√106 = (32080051 + 3115890√106)^1∕2 = 8010.0001248.
Table XV shows first triples for the Pell equation x² − 113y² = ±1 using the first least solution (1204353,113296,1) and the multiplicand R₃ = 776 + 73√113 = (1204353 + 113296√113)^1∕2 = 1552.000644.

Table XIV D=106
x	y	z
4005	389	-1
32080051	3115890	1
256961212515	24958279289	-1

Table XV D=113
x	y	z
776	73	-1
1204353	113296	1
1869156632	175835465	-1

This concludes Part III. To see two new methods for generating Pellian triples from known sequences go to Part IV and Part V.
Go back to Part IIIA.

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The Pellian Equation x2 −Dy2 = ±1 Continuation (Part IIIB)

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

The Actual Method

Tables of D and Pell (x,y,±1) and (x,y,−1) Triples

The Pellian Equation x² −Dy² = ±1 Continuation (Part IIIB)