My Math Forum # of Pythagorean n-tuples

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 October 3rd, 2017, 10:53 AM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 317 Thanks: 26 Math Focus: Number theory # of Pythagorean n-tuples For n > 1, what count of Pythagorean n-tuples tends to be greatest?
 October 3rd, 2017, 12:20 PM #2 Global Moderator   Joined: May 2007 Posts: 6,448 Thanks: 565 What is being counting? Number of triplets is infinite.
 October 3rd, 2017, 04:19 PM #3 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 317 Thanks: 26 Math Focus: Number theory I guess they all belong to sets that are uncountable, i.e. relatively the same size?
 October 3rd, 2017, 06:59 PM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 309 Thanks: 160 Math Focus: Dynamical systems, analytic function theory, numerics An interesting fact is that during the course of the deriving the generating relations for all pythagorean triples one proves a bijection between triples and ration numbers. Thus, they are countably infinite. The basic idea is a fun enough exercise. Start with $x^2 + y^2 = z^2$. Divide through by $z^2$ so that a triple corresponds to a point on the unit circle. Now find any 1 solution to this equation and realize that any line with a rational slope passing through your solution will intersect the circle once more in another solution. From here one can get explicit generating relations for generating all pythagorean triples. Thanks from greg1313 and Loren
 October 3rd, 2017, 08:47 PM #5 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 $\displaystyle {x_1}^2+{x_2}^2+...+{x_n}^2=y^2$ $\displaystyle x_1=2{p_1}s$ $\displaystyle x_2=2{p_2}s$ ..... ...... ...... $\displaystyle x_{n-1}=2{p_{n-1}}s$ $\displaystyle x_n={p_1}^2+{p_2}^2+....{p_{n-1}}^2-s^2$ $\displaystyle y={p_1}^2+{p_2}^2+....+{p_{n-1}}^2+s^2$ Last edited by Individ; October 3rd, 2017 at 08:50 PM.

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