Thread: Pythagorean 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?

Originally Posted by WinkyLozza

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?

1. You easily can check that this formula satisfies the Pythagorean theorem.

2. You have to observe the constraint

3. I don't know what you mean by an "original formula"?

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)

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.)