Jump to content

User talk:Gwaihir

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia

Welcome from Redwolf24

[edit]

Welcome!

Hello, and welcome to Wikipedia. Thank you for your contributions. I hope you like the place and decide to stay. We as a community are glad to have you and thank you for creating a user account! Here are a few good links for newcomers:

I hope you enjoy editing here and being a Wikipedian! By the way, please be sure to sign your name on Talk and vote pages using four tildes (~~~~) to produce your name and the current date, or three tildes (~~~) for just your name. If you have any questions, see the help pages, add a question to the village pump or ask me on my Talk page. Again, welcome!

Redwolf24 9 July 2005 07:19 (UTC)

P.S. I like messages :-P


Your clue was the final piece of the puzzle I needed to nail this problem. In fact I had hit upon the idea of as a criteria for generating the triples for the original equation. But I was unable to prove that multiples of basic Pythagorean triples where were all that's needed. Essentially, my mental block was on the fact that either the numbers in a Pythagorean triple are all coprime with each other or all three share the same common factor. I kept on trying to find the case where two number out of the three had a common factor that was not divisible by the third, turned out this is impossible, because for a Pythagorean triple :

Assume that has no common factor with or , but and have a greatest common factor, , that is greater than 1. Then

Clearly, is also divisible by , which is a contradiction with the original assumption that has no common factor with or .

Assume that has no common factor with or , but and have a greatest common factor, , that is greater than 1. Then

Clearly, is also divisible by , which is a contradiction with the original assumption that has no common factor with or .

Symmetrically, it could be shown that it is impossible for and to share a greater than 1 common factor that is not divisible by .

Hence, for Pythagorean triples, either all three numbers have a greater-than-one common factor or they are pairwise coprime, that means and and all at the same time! There are no other situations possible in terms of common factors.

With this proof, al[[Image:ong with gwaihir's clue, it is not hard to figure out the following possible values for :

15, 30, 45, 60, 75, 90              for the (3, 4, 5) case
65 for the (5, 12, 13) case

Thus, the answer to the original question (the sum of all such ) is: 350.

Hurrrrrrrrrrrah]], thank you so much for your help! I award you this barnstar:

129.97.252.63 05:02, 21 February 2006 (UTC)[reply]

Thank you, I'm glad I could help you. (But see my comments on WP:RD/Math.)--gwaihir 10:42, 21 February 2006 (UTC)[reply]
Waaaaaaaaaaaaaah, I'm stupid, I'm stupid, yeah (4, 3, 5) should be counted, whereas (12, 5, 13) would pop you over 100. 129.97.252.63 21:07, 21 February 2006 (UTC)[reply]
I added up the numbers again and it's still 350. That is assuming my incomplete table of values. 129.97.252.63 21:07, 21 February 2006 (UTC)[reply]
No, the negative b's and c's need not to be counted, since all I care about here are 's such that has integer/diophantine solutions. so the overlap of a=60 also is included only once. 129.97.252.63 21:07, 21 February 2006 (UTC)[reply]

thanks for the translations

[edit]

I appended them to the German headers. --Jtir 20:51, 9 October 2006 (UTC)[reply]