(12C) Pythagorean Triple
09-18-2019, 11:37 AM (This post was last modified: 09-19-2019 12:48 AM by Gamo.)
Post: #1 Gamo Senior Member Posts: 674 Joined: Dec 2016
(12C) Pythagorean Triple
Program to generate a Pythagorean Triples based on this formula

a = m^2 - n^2 , b = 2mn , c = m^2 + n^2

for m > n with m and n co-prime and not both odd

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Procedure: FIX 0

Your choice of integer m > n

m [ENTER] n [R/S] display answer a ... b ... c [R/S] add one on both m and n and display next result.

To start over with new pair of m and n f [PRGM]
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example: f [PRGM]

m = 2 , n = 1

2 [ENTER] 1 display 3 .... 4 .... 5 [R/S] 5 ... 12 ... 13

To review result use [R↓] [R↓] [R↓]

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Program:
Code:
 01 STO 2  // n 02 R↓ 03 STO 1  // m 04 RCL 1 05 RCL 2 06  - 07 LSTx 08 X<>Y 09 ENTER 10  x ------------- 11 √x 12 X=0 13 GTO 15 14 GTO 06 15 X<>Y  // GCD routine 16  1 17  - 18 X=0  // Test for a co-prime 19 GTO 24 20  1 ----------- 21 STO+1 22 STO+2 23 GTO 04 24 RCL 1 25  2 26  ÷ 27 FRAC 28 X=0  // Test for odd or even integer 29 GTO 39 30 RCL 2 ----------- 31  2 32  ÷ 33 FRAC 34 X=0  // Test to make sure not both odd integer 35 GTO 39 36  1 37 STO+1 38 GTO 04 39 RCL 1 40 ENTER ----------- 41  x 42 RCL 2 43 ENTER  44  x 45  + 46 STO 3 47 RCL 1 48 RCL 2 49  x 50  2 ---------- 51  x 52 STO 4 53 RCL 1 54 ENTER 55  x 56 RCL 2 57 ENTER  58  x 59  - 60  0 ---------- 61 X<>Y 62 PSE  // display a 63 RCL 4 64 PSE // display b 65 RCL 3  // display c 66 R/S 67  1 68 STO+1 69 STO+2 70 GTO 04  // add one and start over

Gamo
09-20-2019, 04:33 PM
Post: #2
 Albert Chan Senior Member Posts: 1,461 Joined: Jul 2018
RE: (12C) Pythagorean Triple
If circle x² + y² = c = m² + n², we know P = (m,n) is on the circle.

We can get rational parametizations of the circle, using this rational point.

Let line that pass thru P, y = (x-m)t + n.
To get its intersection of the other point, Q, substitute it into circle equation:

x² + ((x-m)t + n)² = m² + n²

(1+t²) x² + (2nt - 2mt²) x + (-m² - 2mnt + m²t²) = 0

Product of roots, xP xQ = m xQ = (-m² - 2mnt + m²t²) / (1+t²)

xQ = (-m - 2nt + mt²) / (1+t²)

Thus, rational parametized points on the circle is

$$\Large Q = \left( {-m - 2nt + mt² \over 1+t²} , {n - 2mt - nt² \over 1+t²} \right)$$

If circle is unit circle, and we pick P=(-1, 0), we get trig substitution form, t=tan(½θ)

$$\Large Q = (\cos θ, \sin θ) = \left( {1 - t² \over 1+t²}, {2t \over 1+t²} \right)$$

Example, with P=(-5,0), and for positive ts we get Qs:

$$\large (0,5), (-3,4), (-4,3), ({-75 \over 17}, {40 \over 17}), ({-60 \over 13}, {25 \over 13}), ({-175 \over 37}, {60 \over 37}), ({-24 \over 5}, {7 \over 5}), ({-63 \over 13}, {16 \over 13}) ...$$

ref: "The Irrationals", appendix B, by Julian Havil
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