Generating Primitive Pythagorean Triples

Summary:

Using Linear Algebra

Generating Primitive Pythagorean Triples

jonas2 02 Dec 2008 17:34

Taken from the Wolfram website. I am pretty sure this method wasn't covered but I can't always be sure…

If $<a_0, b_0, c_0>$ is a Primitive Pythagorean Triple, then $<a_0, b_0, c_0>U_i$ generates a new primitive triple $<a_i, b_i, c_i>$ where

$U_1 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_2 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_3 = \left| \begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right|$.

Let's exam a simple example. We were told that $<3, 4, 5>$ is a primitive triple. So…

$<3 ,4, 5>U_1 = <5, 12, 13> ,$ $<3, 4, 5>U_2 = <21, 20, 29> ,$ $<3, 4, 5>U_3 = <15, 8, 17> ,$

…which can easily be verified to be valid!

Last edited on 03 Dec 2008 05:09 by jonas2 Show more

Re: Generating Primitive Pythagorean Triples

Greg Gifford 05 Dec 2008 04:41

There's also another way to generate primitive triples that I don't think was discussed in class, although it isn't linear algebra. If you let $a = 2k+1$ (any odd number), then $b = (a^2-1)/2$ and $c = (a^2+1)/2$. To prove it:

$a^2+b^2 = c^2$
$(2k+1)^2+(((2k+1)^2-1)/2)^2 = (((2k+1)^2+1)/2)^2$
$(2k+1)^2+((4k^2+4k+1-1)/2)^2 = ((4k^2+4k+1+1)/2)^2$
$(2k+1)^2+(2k^2+2k)^2 = (2k^2+2k+1)^2$
$4k^2+4k+1+4k^4+8k^3+4k^2 = 4k^4+8k^3+8k^2+4k+1$
$4k^4+8k^3+8k^2+4k+1 = 4k^4+8k^3+8k^2+4k+1$

Re: Generating Primitive Pythagorean Triples

eguzman2 07 Dec 2008 06:55

Here is something that I wanted to tell you about in class but didn't not had time:

Plato’s Formula:
Let m be contained the set of integers with m > 1. Then we can generate a Pythagorean triple by letting a = 2m, b = m^2 – 1, c = m^2 + 1.

Ex. Let m = 2, so we have that a = 2(2) = 4, b = 4 – 1 = 3, c = 4 + 1 = 5.
Then 4^2 + 3^2 = 5^2, which is true.
The formula generates finitely many triple but not all Pythagorean triples.

Formula for generating all Pythagorean triples

Consider the Pythagorean triple x – y – z where (x,y,z) = d. Let x = du,
y = dv, and z = dw where (u,v,w) = 1. So we have u^2 + v^2 = w^2, thus
u – v – w is a triple. Then we can say that every Pythagorean triple is a multiple of primitive triple.

Ex. Let x = 6, y = 8, z = 10. We have that 6^2 + 8^2 = 10^2. Then (6,8,10) = 2. We have that:
x = 2(3) = 6
y = 2(4) = 8
z = 2(5) = 10

This tells that 3 – 4 – 5 is a Pythagorean triple, which we know is true.

Using this idea in the equations for generating infinitely many primitive triples that Dan showed we can generate all triples by adding a parameter d. So we have:

a = d*(m^2 – n^2), b = d*(2mn), c = d*(m^2 + n^2) where m, n, d are contained in the set of integers with m>n>0 and d positive, n or m is odd and (m,n) = 1.

Conclusion:
The formula generates all Pythagorean triples but not uniquely.

Last edited on 07 Dec 2008 06:55 by eguzman2 Show more