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

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,15,1), where seven D values (176-904) 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, 3, 176, 223, 227, 280, 799, 896, 904, 1007, 1872, 2019, 2031, 2184, 3395, 3592, 3608, 3811, 5368, 5615, 5635, 5888, 7791,...

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 = 3
P(n) =
F_2n+1 = F_2n + 173(2n-1) F_2n+2 = F_2n+1 + 104n
P(m) = (m(324m − 1)), (m(324m + 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 = 3 and the counter n set to 1. Thus, according to the second line F₃ = 176 and F₄ = 280 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:

n and m are both initially set to 1 and incremented by 1
173 is the difference between 3 and 176, and 104 is the difference between 176 and 280
Each pair of P(n) and P(m) employs the same respective n and m

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 15.

Table D
n	1	2	3	4	5	6	7	8	9	10	11	12	13	14
D(n)	3	176	280	799	1007	1872	2184	3395	3811	5368	5888	7791	8415	10664
x	26	199	251	424	476	649	701	874	926	1099	1151	1324	1376	1549

D(m)	223	227	896	904	2019	2031	3592	3608	5615	5635	8088	8112	11011	11039
x	224	226	449	451	674	676	899	901	1124	1126	1349	1351	1574	1576

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 + 3√D)² ∕ ¹⁄_2.5n₁ = x + 15√D
R(m)_D = (15m₁ + √D)² ∕ ¹⁄_7.5m₁ = x + 15√D

where:

n₁ are the integers: 35n + 5(2n ∓ 1) starting at n = 0
that is n is repeated twice in order to satisfy first the −1 expression then the +1 expression
where, the first n₁ value −5 is thrown out
the values of adjacent xs in a pair is (225(2n − 1) − 173)/2), (225(2n − 1) + 173)/2)
D are the values from the above sequence starting at 3.

m₁ are consecutive integers: 1, 2, 3, 4, 5, ...
the values of adjacent xs in a pair is (225m₁ − 1), (225m₁ + 1)
D are the values from the above sequence starting at 223.

where n₁ pairs (0,1), (1,2), (2,3), (3,4),(4,5), etc. correspond to the pairing of the R(n)_D equations as shown in Table I.

Pell Equal Expressions for (x,15,1) Triples

Note: All the (5n₁ + √D)² ∕ ¹⁄_2.5n₁ and (15m₁ + √D)² ∕ ¹⁄_7.5m₁ are present in pairs. In addition, a D(n) pair is shown first followed by a D(m) pair.

Table I
n	2n∓1	Pell Equation	Equal Expressions
0	1	x² − 3y² = 1	R₃ = (5 + 3√3)² ∕2 = 26 + 15√3
1	1	x² − 176y² = 1	R₁₇₆ = (40 + 3√176)² ∕16 = 199 + 15√176
		x² − 223y² = 1	R₂₂₃ = (15 + √223)² ∕2 = 224 + 15√223
		x² − 227y² = 1	R₂₂₇ = (15 + √227)² ∕2 = 226 + 15√227
1	3	x² − 280y² = 1	R₂₈₀ = (50 + 3√280)² ∕20 = 251 + 15√280
2	3	x² − 799y² = 1	R₇₉₉ = (85 + 3√799)² ∕34 = 424 + 15√799
		x² − 896y² = 1	R₈₉₆ = (30 + √896)² ∕4 = 449 + 15√896
		x² − 904y² = 1	R₉₀₄ = (30 + √904)² ∕4 = 451 + 15√904
2	5	x² − 1007y² = 1	R₁₀₀₇ = (95 + 3√1007)² ∕38 = 476 + 15√1007
3	5	x² − 1872y² = 1	R₁₈₇₂ = (130 + 3√1872)² ∕52 = 649 + 15√1872
		x² − 2019y² = 1	R₂₀₁₉ = (45 + √2019)² ∕6 = 674 + 15√2019
		x² − 2031y² = 1	R₂₀₃₁ = (45 + √2031)² ∕6 = 676 + 15√2031

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

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

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

Pell Equal Expressions for (x,15,1) Triples

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