User talk:Gwaihir

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

$$c^2 = a^2 + b^2 = (a'd)^2 + (b'd)^2 = a'^2d^2 + b'^2d^2 = d^2(a'^2 + b'^2)$$

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 [[User Talk:Gwaihir|gwaihir]]'s clue, it is not hard to figure out the following possible values for $$a$$:

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:

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)

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