Sequences of Pythagorean Multiples (Part XIV)

Use of the Diophantine Equation x2 ± D(∑ yj2) = z2 for Generating the Sequences

In Part XII Diophantine triples using the Diophantine equation x2 + D(∑ yj2) = z2 was used to produce four examples. This section will show how to create consecutive triples of the Pythagorean equation to generate sequences where the a, b and c are integral numbers. In addition,a new modified arithmetic progression method and a new random access method to generate the triples will be shown. We first produce all triples where all the aj are equal and where the subscript j increases consecutively from 1 to k. This is shown in Table I where for example all the ajs are equal to 3.

Table I (D = 1, n= 1)
x y1y2y3...ykz
a1a2a3...ak D(a12 + a22 + a32 + ... + ak2) − n2 2na12na22na3 D(a12 + a22 + a32 + ... + ak2) + n2
300(9) − 1 = 860010
330(9 + 9) − 1 = 1766019
333(9 + 9 + 9) − 1 = 2666628
333...3(9 + 9 + 9 + ...) − 1 = 9k − 1666...69k + 1

When the calculations are done for k = 1 to 16, the following results are obtained and listed in Tables IIa and IIb, where the δy2 values are used to avoid non-integral values of y. In addition, the δ1 between adjacent x and z values is 9 which may also be obtained from the arithmetic progression formula using 8 as the initial number, ai, and 9 as the common difference, d:

ak = ai + d(k − 1))     (A1)

Inspection of the y2 column shows, that only the bold/italized values give y integral values of 6, 12, 18 and 24 and where their common difference is 6. We can extract these particular italized triples from these two tables and generate a sequence of integral Pythagorean triples (Table III). Alternatively, the xs can be obtained by multiplying (aj)2 (in this case 9) with multiples of m2 = 1,4,9,16,... then subtracting 1 as exhibited in Table I.

Table IIa (δy2 =36)
δ1x y2 zδ1
83610
99
177219
99
2610828
99
3514437
99
4418046
99
5321655
99
6225264
99
7128873
99
Table IIb (δy2=36)
δ1x y2 zδ1
8032482
99
8936091
99
98396100
99
107432109
99
116468118
99
125504127
99
134540136
99
143576145

On inspection we see that the common difference between adjacent x and z values is δ1 and that the difference between adjacent δ1s is δδ = 18. Moreover, when we get to the end of Table II we can continue generating new x and z2 values by using the various delta (δ) differences.

Table III (δy=6)
δδδ1x y zδ1δδ
8610
2727
1835123718
4545
1880188218
6363
181432414518
8181
182243022618
9999
183233632518
117117
184404244218
135135
185754857718
153153

The x and z values may alternatively be obtained using equation (A2) as long as z is not of the form non-integral z, since this must require a different way of generating the common difference.

bk+1 = bk + (ai + (k − 1)d)    (A2)

where ai is the initial δ1 value, d is the common difference δδ and bk are the variables x or z. The bk+1, thus, obtained in the first step becomes the bk of the next step. On the other hand, the values for y are obtained using the regular arithmetic progression formula (A1).

We place the initial numbers for bks of 8 and 10 in equation (A2) and perform the calculations for the various bks in Tables IVx and IVz. We see that these bk values are in agreement with those of Table III.

Table IVx (d = δδ = 18)
bk+1 = bk + (ai + (k − 1)d)
b2 = 8 + (27 + 0(18)) = 35
b3 = 35 + (27 + 1(18)) = 80
b4 = 80 + (27 + 2(18)) = 143
b5 = 143 + (27 + 3(18)) = 224
b6 = 224 + (27 + 4(18)) = 323
b7 = 323 + (27 + 5(18)) = 440
b8 = 440 + (27 + 6(18)) = 575
b9 = 575 + (27 + 7(18)) = 728
Table IVz (d = δδ = 18)
bk+1 = bk + (ai + (k − 1)d)
b2 = 10 + (27 + 0(18)) = 37
b3 = 37 + (27 + 1(18)) = 82
b4 = 82 + (27 + 2(18)) = 145
b5 = 145 + (27 + 3(18)) = 226
b6 = 226 + (27 + 4(18)) = 325
b7 = 325 + (27 + 5(18)) = 442
b8 = 442 + (27 + 6(18)) = 577
b9 = 577 + (27 + 7(18)) = 730

Generating Pythagorean Multiples Via Random Access

Though consecutive Pythagorean triples can be obtained via this manner, a new way of generating the x and z values has been produced (Part XIX). Since each triplet is k2 distance from each other, the equations are modified to k2xi + Δ ∕2(k2−1) for the value of x and k2xi + Δ ∕2(k2+1) for the value of z. Δ corresponds to the difference zx in Table III and xi corresponds to the initial value of x in the table and k can be any value greater than zero. Thus, for example:

Δ = 2, k = 8 and xi = 8
x: 64 × 8 + 63 = 575
z: 64 × 8 + 65 = 577

Identical to the last b8 values of Tables IIIx and IIIz. If we take k = 25 we get:

Δ = 2, k = 25 and xi = 8
x: 625 × 8 + 624 = 5624
z: 625 × 8 + 626 = 5626
z2x2 = 56262 − 56242 = 1502

Thus, we can access randomly Pythagorean triples of this type via the use of these equations.

This concludes Part XIV. The next section (Part XV) will show just the tables of fully integral triples for aj of 5 and 7 obtained by the above method.

Go back to Part XIII. Go back to homepage.


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