Hey, I found a formula to find pythagorean triads with ease

c - k = b

a = √(2.k.c - k^2)

where:

c is hypotenuse

a and b are sides

k is any number (constant)

Is this formula original? If so, does this work?

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- Nov 19th 2010, 10:43 PMWinkyLozzaPythagorean Triples Formula
Hey, I found a formula to find pythagorean triads with ease

c - k = b

a = √(2.k.c - k^2)

where:

c is hypotenuse

a and b are sides

k is any number (constant)

Is this formula original? If so, does this work? - Nov 19th 2010, 11:57 PMearboth
- Nov 20th 2010, 02:45 PMWinkyLozza
Well, I don't know if someone has found it before, but it took me a whole half an hour to figure out.

Yes, that constraint was common logic for me, so I hadn't thought of including it.

Thanks

PS:

Here's the proof:

I figured you found certain pythag triples by finding that the difference between two adjacent numbers, eg 12 and 13, if the difference's root is an integer:

(c^2 - (c - 1)^2) = a^2

where c - 1 = b

I could then simplify the above

c^2 - c^2 + 2c - 1 = a^2

a^2 = 2c - 1

a = √(2c - 1)

I then tried using the difference for two numbers of difference 2, where the difference of their squares is an integer:

(c^2 - (c - 1)^2) + ((c - 1)^2 - (c - 2)^2) = a ^ 2

where c - 2 = b

Simplify:

c^2 - c^2 + 2c - 1 + c^2 - 2c + 1 - c^2 + 4c - 4 = a^2

a^2 = 4c - 4

a = √(4c - 4)

To save the typing, the result for the two numbers with difference three is:

a = √(6c - 9)

where c - 3 = b

Wait, there's a pattern...

a = √(2(c - 0.5))

a = √(4(c - 1))

a = √(6(c - 1.5))

a = √(2x(c - x/2))

a = √(2xc - x^2) - Nov 20th 2010, 06:49 PMroninpro
Euclid beat you to it.

Pythagorean triple - Wikipedia, the free encyclopedia

(And actually, I think that the Babylonians may have known about the formula as well. You may need to check up on that, though.)