User talk:Gwaihir

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[edit] Welcome from Redwolf24

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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 d\mid ab as a criteria for generating the triples for the original equation. But I was unable to prove that multiples of basic Pythagorean triples where a'd'\leq100 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 (a,b,c):

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

c2 = a2 + b2 = (a'd)2 + (b'd)2 = a'2d2 + b'2d2 = d2(a'2 + b'2)

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

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

c^2 = a^2 + b^2 \Rightarrow a^2 = c^2 - b^2 = (c'd)^2 + (b'd)^2 = c'^2d^2 + b'^2d^2 = d^2(c'^2 + b'^2)

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

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

Hence, for Pythagorean triples, either all three numbers have a greater-than-one common factor or they are pairwise coprime, that means gcd(a,b) = 1 and gcd(b,c) = 1 and gcd(c,a) = 1 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 a:

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 a) is: 350.

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

129.97.252.63 05:02, 21 February 2006 (UTC)

Thank you, I'm glad I could help you. (But see my comments on WP:RD/Math.)--gwaihir 10:42, 21 February 2006 (UTC)
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)
I added up the numbers again and it's still 350. That is assuming my incomplete table of a values. 129.97.252.63 21:07, 21 February 2006 (UTC)
No, the negative b's and c's need not to be counted, since all I care about here are a's such that \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} 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)

[edit] thanks for the translations

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

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