It's that Law of Gravity that's really keeping me down...
By the way, I'm aware that Rich Bond isn't really a money guy. But c'mon! It's just too good a name not to use.
Previous episodes of The 0'Factor: 01, 02, 03, 04, 05, 06, 07, 08, 09.
By the way, I'm aware that Rich Bond isn't really a money guy. But c'mon! It's just too good a name not to use.
Previous episodes of The 0'Factor: 01, 02, 03, 04, 05, 06, 07, 08, 09.
For more on Henry's and Ian's adventures, click on the following links:
Through a Window Darkly: Part I,II,III, IV, V, VI, TVH, TUC.
P_{1} = 1, Q_{1} = 0, Q_{n+1} = P_{n}, P_{n+1} = 2 X P_{n} + Q_{n} To follow my method, calculate 1 + Q/P as an improper fraction. As in my prior post, this will give you a triple in the form (a+b)/c, where b = a + 1. This method was "easier" for me than what Kraitchik (1953) wrote, but it required more iterations to get each Pythagorean Triple. Kraitchik offers a shortcut, which is essentially: a = 2PQ and b = P^{2} - Q^{2}, which yields numbers such that | a - b | = 1 For example: a = 2(2)(1) = 4, b = (2)^{2} - (1)^{2} = 3, c = 5 a = 2(5)(2) = 20, b = (5)^{2} - (2)^{2} = 21, c = 29 a = 2(12)(5) = 120, b = (12)^{2} - (5)^{2} = 119, c = 169 etc. |
Followed by: 99/70 and 239/169 | If you look at the fractions with the odd denominators, you have a Pythagorean Triple in the form of (a+b)/c, where a and b are consecutive numbers (i.e., b = a + 1) 7/5 gives you 3, 4, 5. 41/29 gives you 20, 21, 29 239/169 gives you 119, 120, 169 |
b = (1/8)a^{2} - 2 | c = (1/8)a^{2} + 2 |
a^{2} + b^{2} = (b + 2)^{2} a^{2} + b^{2} = b^{2} + 4b + 4 a^{2} = 4b + 4 a^{2} = 4(b + 1) | c = b + 2 c = (1/4)a^{2} - 1 + 2 c = (1/4)a^{2} + 1 |
a^{2} + b^{2} = (b + 1)^{2} a^{2} + b^{2} = b^{2} + 2b + 1 a^{2} = 2b + 1 | c = b + 1 c = (1/2)(a^{2} - 1) + 1 c = (1/2)(a^{2} + 1) |