Finding the Diophantine Equation x2 ±Dy2 = z2 (Part IX)

A Method for Recovering k Given the Triple (x,y,z)

Given the triple (x,y,z) can one find k, the position on the table where that triple is to be found for the Diophantine equation x2 ± Dy2 = ±z2, i.e, work backwards. Normally one generates a set of equations that is used to generate the triples, x, y and z. These triples are then tabulated into a table having the following table heading format:

δ1x y z δ2
k Dk2 + n mk Dk2 - n

where k is a counter starting at zero, the numeral D is the coefficient of y and the (n, m) are values taken from the table below so that whenever n = j2 then m = 2j for all j > 0:

Table of n & m values
n149162536496481100...
m2468101214161820...

Two sets of triples were chosen from Table 91, page 254 of Recreations in the Theory of Numbers by Albert H. Beiler (1966) for the least solution of the Pell equation, x2 − Dy2 = 1. The method on this page generates the k position not only for this equation but also for x2 − Dy2 = −1. Thus the method is applicable to both triples (x,y,±z) where y is an even number. Odd number y requires different equations as shown in Part I/IIA and switching the a and c columns for (a,b,−c) triples.

The Pell equations x2 − 30y2 = ±1

We are given (11,2,1) as the least solution for the Pell equation x2 − 30y2 = 1. We start by setting the value for n in column x in the table above to j2 and then setting the equation equal to 11j (where j is a multiple of 11) and rearranging to:

x = 30k2 + j2 = 11j
j2 − 11j + 30k2 = 0

Using the quadratic equation to find the value of j:

j = (11 ± √ 112 − 4×30k2 )/2 = (11 ± √ 121 − 120k2 )/2

Setting k = 1:

j = (11 ±1)/2 giving two values for j = 5,6

The Solution for x2 − 30y2 = 1

Plugging the value of 5 into x, y and z gives:

x = 30 + 25 = 55, y = 10 and z = 30 − 25 = 5

Thus (55,10,5) is found at k = 1 of the table where the equations are x = 30k2 + 25, y = 10k and z = 30k2 − 25.

The Solution for x2 − 30y2 = −1

And plugging the value of 6 into x, y and z gives:

x = 30 + 36 = 66, y = 12 and z = 30 − 36 = −6

Thus (66,12,−6) is found at k = 1 of the table where the equations are x = 30k2 + 36, y = 12k and z = 30k2 − 36.

The Pell equations x2 − 13y2 = ±1

We are given (649,180,1) as the least solution for the Pell equation x2 − 13y2 = 1. We start by setting the value for n in column x in the table above to j2 and then setting the equation equal to 11j (where j is a multiple of 11) and rearranging to:

x = 13k2 + j2 = 649j
j2 − 649j + 13k2 = 0

Using the quadratic equation to find the value of j:

j = (649 ± √ 6492 − 4×13k2 )/2 = (649 ± √ 421201 − 52k2 )/2 = (649 ± √ 421201 − 421200 )/2

where 52k2 = 421200 in order for the difference under the square root to equal to one and:

4k2 = 421200, k2 = 8100 and k = 90
j = (649 ±1)/2 giving two values for j = 324,325

The Solution for x2 − 13y2 = 1

Plugging the value of 324 into x, y and z gives:

x = 13(8100) + 3242 = 649(324) = 210276
y = 90(648) = 58320
z = 13(8100) − 3242 = 1(324)

Thus (649,180,1) is found at k = 90 of the table where the equations are x = 13k2 + 3242, y = 648k and z = 13k2 − 3242. However, 324 is the square of 18 so that using 324 instead of 3242 and k = 5 instead of k = 90 gives the simpler result:

x = 13(25) + 324 = 649
y = 36(5) = 180
z = 13(25) − 324 = 1

Thus (649,180,1) is found at k = 5 of the table where the equations are x = 13k2 + 324, y = 36k and z = 13k2 − 324.

The Solution for x2 − 13y2 = −1

And plugging the value of 325 into x, y and z gives:

x = 13(8100) + 3252 = 649(325) = 210925
y = 90(650) = 58500
z = 13(8100) − 3252 = −1(325)

Thus (649,180,−1) is found at k = 90 of the table where the equations are x = 13k2 + 3252, y = 650k and z = 13k2 − 3252. Because the square root of 325 is not an integer giving a simpler answer is not possible.

This concludes Part VI.

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Copyright © 2020 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com