Is there a deterministic method for generating all of the Pythagorean triples {a, b, c} for a given hypotenuse c?
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The answer to your question as stated is trivially "yes": Given hypotenuse $c$, check, for each natural $a<c$, whether $c^2-a^2$ is a perfect square. Were you looking for an efficient deterministic algorithm? – dfeuer Nov 06 '13 at 03:03
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@abiessu "Pythagorean triples" are integral by definition. – MJD Nov 06 '13 at 03:06
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Are you familiar with the theorem about which integers can be represented as the sum of two squares? – MJD Nov 06 '13 at 03:08
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Check http://math.stackexchange.com/questions/518856/integral-points-on-a-circle/518944#518944 – Macavity Nov 06 '13 at 03:30