Pyth. Triples Formula

[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?

[dxoigmn]

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.

[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).

[Maddox]

This is not an unknown formula.

[K]

This is not an unknown formula.

The idea behind the encouragement is not that it's an unknown formula, but rather that he came up with it independently.

[CrAzY]

Thanks... I'll name it the Yoni Forumlas because Yoni is nice... You figure one out by your self maddox