Never mind previous post. I assumed something in "proof" of pythagorean question that's not strictly true. Now not sure if it's obvious how to classify all possible pythagorean triplets.
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12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
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I of course generate an infinite series of answers but they don't exhaust possibilities. My inductive step was flawed...12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
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The question is, for each number, does there exist a unique set of pythagorean triplets where this number is the smallest?Be good, and if at first you don't succeed, perhaps failure will be back in fashion soon. -- teh Spamski
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If n = 2 (mod4) then obviously not.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
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Can there be multiple sets for a number though?Be good, and if at first you don't succeed, perhaps failure will be back in fashion soon. -- teh Spamski
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Reasoning as follows:
y^2 - x^2 = n^2
Obviously either x odd and y odd or x even and y even (if not then n^2 must be odd, a contradiction)
If x even and y even then there is a smaller equivalent triplet
Therefore x odd and y odd. Let y = x+2i => y^2 -x^2 = 4ix + 4i^2 = 4(ix + i^2). However, for x odd ix + i^2 is even for all i (if i odd then you are adding an odd to an odd; if i even then an even to an even) so y^2 - x^2 is divisible by 8. However n = 2(mod 4) cannot give n^2 = 0(mod 8), so x odd and y odd impossible12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
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Originally posted by Snowflake
The question is, for each number, does there exist a unique set of pythagorean triplets where this number is the smallest?12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
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Originally posted by KrazyHorse
In fact, it's quite easy to show that up to a scale factor all pythagorean triplets are equivalent to a = (2n + 1), b = (2n^2 + 2n), c = (2n^2 + 2n + 1)
It is the case up to a scale factor, as I've said. For all intents and purposes 3 4 5, 5 12 13, 7 24 25 etc. are the only pythagorean triplets. All others are direct multiples of these.
I'm surprised that no one else here knows (or knew until snwflake went hunting) of the very simple formula for generating all of the triplets: a=u^2-v^2, b=2uv, c=u^2+v^2 for integers u>v>0. That's how I came up with my counterexample 20-21-29; I just tested values of u and v until fidning one (5 and 2) that worked.
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Originally posted by Snowflake
Confidence intervals will not give you much more info in this case, I believe. He has good statistical inferences (low t-score, high r square, etc.), the problem is that with only a few observations in the sample, how much could we trust those statistical inferences.
I still havent found a good answer so I guess I'll throw that point back to the journal to decide.We need seperate human-only games for MP/PBEM that dont include the over-simplifications required to have a good AI
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Give me a natural number, and I'll give you all the possible pythagorean triplets where this number is the shortest side.Last edited by Snowflake; January 14, 2005, 10:53.Be good, and if at first you don't succeed, perhaps failure will be back in fashion soon. -- teh Spamski
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Originally posted by SpencerH
In theory t-test is applicable to small numbers but the problem is that its difficult to decide whether the data violate the t-test assumptions. IIRC violating the assumptions reduces the sensitivity of the test but doesnt necessarily invalidate results where the null hypothesis is rejected.
I still havent found a good answer so I guess I'll throw that point back to the journal to decide.Be good, and if at first you don't succeed, perhaps failure will be back in fashion soon. -- teh Spamski
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