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January 19th, 2014, 07:53 AM   #11
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Re: Formula for all and only primitive Pythagorean triples.

Quote:
Originally Posted by complicatemodulus
As probably already written here I found more interesting that:

Given an odd integer A there are ALWAYS two integer B and C that satisfy: A^2 = C^2-B^2

If you keep:

A, B= (A^2-1)/2, C= (A^2+1)/2 so you can see all triplets sorted by A odds.

To obtain triple for ALL integers you've just to ADD the even 2*A values where A is what above.

Thanks
Ciao
S.
Gonna say same thing :P Learnt it very young from murderous maths
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January 19th, 2014, 09:54 PM   #12
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Re: Formula for all and only primitive Pythagorean triples.

Very sorry I rewrite with more care:

A1) Given an odd integer A, there are ALWAYS two integers B and C that satisfy: A^2 = C^2-B^2

If you keep:

A, B= (A^2-1)/2, C= (A^2+1)/2 so you can see all triplets sorted by A odds.

A2) To obtain triplets for ALL even integers you've to keep for A even:

A, B= [(A/2)^2 ]-1, C= [(A/2)^2 ]+1 so you can see all triplets sorted by A even.

With this 2 formulas you will obtain a triplet for any A from A=1 to infinite.

Thanks & sorry
Ciao
S.
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