Valhalla Legends Forums Archive | Yoni's Math Forum | Pyth. Triples Formula

AuthorMessageTime
CrAzY
I believe I've found an undiscovered formula to generate pythagorean triples. Or should I say, 2...

Given Int(A)

If A is Odd, Then B = (A^2-1)/2; C = B + 1
If A is Even, Then B = ((A^2)/2)-1; C = B + 2

Made a program for it too,
http://addpower.org/TrippleCalculator.exe

Yawn... This is what happens when you get bored while taking Geometry 6 1/2 hours a day durring the summer. :( Guess good things come out of it too?
August 7, 2004, 3:13 AM
kamakazie
Doesn't look complete. Can you generate: A = 15, B = 8, C = 17. Your formula assumes C = B + 1 OR C = B + 2 depending on A, which isn't necessarily true.
August 7, 2004, 4:16 AM
Yoni
Very nice!

For any odd integer A, let B = (A^2 - 1) / 2, then:
A^2 + B^2 = A^2 + (A^4 - 2A^2 + 1) / 4 = (A^4 + 2A^2 + 1) / 4 = ((A^2 + 1) / 2) = (B + 1)^2
Hence A,B,B+1 is a pythagorean triple.

For an even integer, your formula is not correct, but you used a correct one in your sample program. It is:
B = (A / 2)^2 - 1

For any even integer A, let B = (A / 2)^2 - 1, then:
A^2 + B^2 = A^2 + A^4 / 16 - A^2 / 2 + 1 = A^4 / 16 + A^2 / 2 + 1 = ((A^2 / 2)^2 + 1)^2 = (B + 2)^2
Hence A,B,B+2 is a pythagorean triple.

In addition to that, for any such A, the integers B and B+2 are coprimes, giving for every integer n, a unique pythagorean triple with n the least element. (*)

Great work! Keep it up. :)

Edit:
A correction. (*) is only true when A != 2 (mod 4).
August 7, 2004, 1:26 PM
Maddox
This is not an unknown formula.
August 24, 2004, 10:14 AM
K
[quote author=Maddox link=board=36;threadid=8059;start=0#msg77135 date=1093342468]
This is not an unknown formula.
[/quote]

The idea behind the encouragement is not that it's an unknown formula, but rather that he came up with it independently.
August 25, 2004, 4:38 AM
CrAzY
Thanks... I'll name it the Yoni Forumlas because Yoni is nice... You figure one out by your self maddox >:(
September 16, 2004, 2:44 AM

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