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June 28th, 2014, 07:11 AM   #1
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All Pythagorean Triples c²=a²+b² here

(2ab)² + (a²-b²)²= (a²+b²)²

ab = n in N

q.e.d.

M_B_S
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June 29th, 2014, 12:36 PM   #2
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This is just Euclid's formula to generate Pythagorean triples.
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July 1st, 2014, 11:39 PM   #3
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Quote:
Originally Posted by eddybob123 View Post
This is just Euclid's formula to generate Pythagorean triples.
Correct!

(k2ab)²+(k(a²-b²))²=(k(a²+b²)²

it comes from

4ab=(a+b)²-(a-b)²

Physics intuitions: A Pythagorean relation for any triangle?
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