The Pellian equation
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.
This article, a continuation of Part XIXB, is in regard to the sequences Sn generated from the the equations of the type N(n)(n+1) + m which have very interesting properties as was shown in Part XVII for 4(n)(n+3) + 5 and in Part IIIA for 4(n)(n+1) + 5. Six other sequences with similar equations are displayed in Part XIXB and these two new sequences in this article are part of this group, in numerical order:
where all the sequences were evaluated at n ≥ 0. The two sequences will be shown to behave like Sn5 and Sn13 of Part XIXB where one of their equivalents is based on the square root of a cubed expression.
All the terms above (the D values) were entered into the Pell calculator program to give values of Qn = +1 for different values of n. It should be noted that the x and the y terms are found at the bottom of column n − 1 just to the left of Qn in the coded examples. We can see that in the two coded examples: for 717 at
Tables I and II show the first 11 D numbers in the two different sequences Sn. Although the values of D equal 69 and 21 in Tables I and II, respectively, they are tabulated even though they do not have the requisite Qn = 1 at n = 13 and 11, respectively, shared with the other D terms in the equations.
D | −3 | 69 | 213 | 429 | 717 | 1077 | 1509 | 2013 | 2589 | 3237 |
---|---|---|---|---|---|---|---|---|---|---|
x | - | 7775 | 194399 | 1524095 | 6998399 | 23522399 | 64393055 | 152409599 | 323606015 | 631605599 |
y | - | 936 | 13320 | 73584 | 261360 | 716760 | 1657656 | 3396960 | 6359904 | 11101320 |
yf | - | 3(312) | 5(2664) | 7(10512) | 9(29040) | 11(65160) | 13(127512) | 15(226464) | 17(374112) | 19(584280) |
D | 21 | 93 | 237 | 453 | 741 | 1101 | 1533 | 2037 | 2613 | 3261 |
---|---|---|---|---|---|---|---|---|---|---|
x | 55 | 12151 | 228151 | 1653751 | 7352695 | 24313015 | 65935351 | 155143351 | 328116151 | 638642935 |
y | 12 | 1260 | 14820 | 77700 | 270108 | 732732 | 1684020 | 3437460 | 6418860 | 11183628 |
yf | 1(12) | 3(420) | 5(2964) | 7(11100) | 9(30012) | 11(66612) | 13(129540) | 15(229164) | 17(377580) | 19(588612) |
where the values of y for Sn-3 and Sn21 are the consecutive odd numbers and D and y are divisible by three.
The Pell equations are set up according to the following format where RD is equal to the equivalent expressions and where n1 is a constant multiplied by a consecutive triangular number and (2n+1) is a consecutive odd number:
where n ≥ 2 for all but 21 and 69.
Tables I and II give the formulas for all 10 Pell equations according to the format listed above with the triangular and odd numbers in blue. Note that the values of n1 are off for R21 and R69.
(Sn-3) Equal Expressions | (Sn21) Equal Expressions |
---|---|
R21 = [(3 + 1√21)2 ∕4]3/2 = 55 + 1(12)√21 | |
R69 = [(25 + 3√69)2 ∕4]3/2 = 7775 + 3(312)√69 | R93 = [(29 + 3√93)2 ∕4]3/2 = 12151 + 3(420)√93 |
R213 = [(73 + 5√213)2 ∕4]3/2 = 19439 + 5(2664)√213 | R237 = [(77 + 5√237)2 ∕4]3/2 = 228151 + 5(2964)√237 |
R429 = [(145 + 7√429)2 ∕4]3/2 = 1524095 + 7(10512)√429 | R453 = [(149 + 7√453)2 ∕4]3/2 = 1653751 + 7(11100)√453 |
R717 = [(241 + 9√717)2 ∕4]3/2 = 6998399 + 9(29040)√717 | R741 = [(245 + 9√741)2 ∕4]3/2 = 7352695 + 9(30012)√741 |
R1077 = [(361 + 11√1077)2 ∕4]3/2 = 23522399 + 11(65160)√1077 | R1101 = [(365 + 11√1101)2 ∕4]3/2 = 24313015 + 11(66612)√1101 |
R1509 = [(505 + 13√1509)2 ∕4]3/2 = 64393055 + 13(127512)√1509 | R1533 = [(509 + 13√1533)2 ∕4]3/2 = 65935351 + 13(129540)√1533 |
R2013 = [(673 + 15√2013)2 ∕4]3/2 = 152409599 + 15(226464)√2013 | R2037 = [(677 + 15√2037)2 ∕4]3/2 = 155143351 + 15(229164)√2037 |
R2589 = [(865 + 17√2589)2 ∕4]3/2 = 323606015 + 17(374112)√2589 | R2613 = [(869 + 17√2613)2 ∕4]3/2 = 328116151 + 17(377580)√2613 |
R3237 = [(1081 + 19√3237)2 ∕4]3/2 = 631605599 + 19(584280)√3237 | R3261 = [(1085 + 19√3261)2 ∕4]3/2 = 638642935 + 19(588612)√3261 |
This concludes Part XIXC.
To go back to Part XIXB.
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