HP Forums

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

-------------------------------------------------
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]
-------------------------------------------------
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↓]

--------------------------------------------------
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

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, x_P x_Q = m x_Q = (-m² - 2mnt + m²t²) / (1+t²)

x_Q = (-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

Gamo

Albert Chan