Welcome from Redwolf24 edit

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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