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 March 7th, 2017, 07:24 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 339 Thanks: 26 Math Focus: Number theory Exclusive Pythagorean singles Do there exist Pythagorean triples that do not share a member with any other triple? Or those that share two? Thanks from agentredlum
 March 7th, 2017, 07:53 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,862 Thanks: 968 they can't possibly be different triples and share two elements Thanks from agentredlum and Loren
 March 8th, 2017, 06:44 AM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Perhaps you are asking about Primitive Pythagorean Triples? If you are, then the answer is YES. You can see using Euclid's Formula which gives all Primitive Pythagorean Triples $(a , b , c) = (m^2 - n^2 , 2mn , m^2 + n^2)$ $(3 , 4 , 5) = (2^2 - 1^2 , 2(2)(1) , 2^2 + 1^2)$ Look at $a = 3$ , there is only one way to generate $3$ in Euclid's Formula (try it!) That fixes $m , n$ so $3$ cannot occur in any other triple. *** On the flipside, look at $c = 5$; there are only two ways to generate $5$ (try it!) and indeed there are only two Primitive Pythagorean Triples that have $5$ as a member. Thanks from topsquark and Loren Last edited by skipjack; March 8th, 2017 at 09:14 AM.
March 8th, 2017, 09:15 AM   #4
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Quote:
 Originally Posted by Loren Do there exist Pythagorean triples that do not share a member with any other triple?
Did you mean a specific value or any value?

March 8th, 2017, 09:52 AM   #5
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Quote:
 Originally Posted by skipjack Did you mean a specific value or any value?
Let's say either, as I am not sure what you mean.

Last edited by Loren; March 8th, 2017 at 10:17 AM.

March 8th, 2017, 10:14 AM   #6
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Quote:
 Originally Posted by agentredlum $(3 , 4 , 5) = (2^2 - 1^2 , 2(2)(1) , 2^2 + 1^2)$ Look at $a = 3$ , there is only one way to generate $3$ in Euclid's Formula (try it!) That fixes $m , n$ so $3$ cannot occur in any other triple.
Keeping short leg < 100, these are the ones that appear exactly once:
3,4,7,8,9,11,16,19,23,27,31,32,43,47,49,59,64,67,7 1,79,81,83

Not 99? Nope. 20-99-101 and 99-4900-4901

Last edited by Denis; March 8th, 2017 at 10:17 AM.

 March 8th, 2017, 05:28 PM #7 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 339 Thanks: 26 Math Focus: Number theory Thanks to Euclid. Greater than his spacetime.
 March 9th, 2017, 04:42 AM #8 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 It is interesting to investigate a number like $15$ and see how many times it can be a member of a Pythagorean Triple (not necessarily primitive). Looking at Euclids Formula we see that there is only one way to generate 15. $(a , b , c) = (m^2 - n^2 , 2mn , m^2 + n^2)$ $\ \ \ \ \$ <-- Euclids Formula $(15 , 112 , 113) = (8^2 - 7^2 , 2(8 )(7) , 8^2 + 7^2)$ $\ \ \ \ \$ <-- Primitive Pythagorean Triple We may be tempted to say $15$ occurs only once as a member. This is not the case however due to the fact that if $(a , b , c )$ is a Pythagorean Triple then so is $( ka , kb , kc )$ for any positive integer $k$ Indeed , member $15$ can occur $2$ more times , genrated by the Primitive Pythagorean Triple $(3 , 4 , 5 )$ When $k = 3$ we generate $(9 , 12 , 15 )$ When $k = 5$ we generate $(15 , 20 , 25 )$ Member $15$ can also be generated by the Primitive Pythagorean Triple $(5 , 12 , 13)$ when $k = 3$ giving $(15 , 36 , 39)$ I found $4$ distinct Pythagorean Triples that have $15$ as a member. Are there more? *** There is a rich underlying structure here that depends on the total number of ways an integer can be written as the sum and/or the difference of $2$ squares. Whether a member is prime or composite (like $15$) also affects the total number of appearances in distinct Pythagorean Triples

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