Generating Primitive Pythagorean Triples
Forum » Ideas / Student Ideas » Generating Primitive Pythagorean Triples
Started by: jonas2jonas2
On: 1228239293|%e %b %Y, %H:%M %Z|agohover
Number of posts: 3
rss icon RSS: New posts
Summary:
Using Linear Algebra
Generating Primitive Pythagorean Triples
jonas2jonas2 1228239293|%e %b %Y, %H:%M %Z|agohover

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 1228280983|%e %b %Y, %H:%M %Z|agohover by jonas2 + show more
unfold Generating Primitive Pythagorean Triples by jonas2jonas2, 1228239293|%e %b %Y, %H:%M %Z|agohover
Re: Generating Primitive Pythagorean Triples
Greg GiffordGreg Gifford 1228452111|%e %b %Y, %H:%M %Z|agohover

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

unfold Re: Generating Primitive Pythagorean Triples by Greg GiffordGreg Gifford, 1228452111|%e %b %Y, %H:%M %Z|agohover
Re: Generating Primitive Pythagorean Triples
eguzman2eguzman2 1228632926|%e %b %Y, %H:%M %Z|agohover

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 1228632952|%e %b %Y, %H:%M %Z|agohover by eguzman2 + show more
unfold Re: Generating Primitive Pythagorean Triples by eguzman2eguzman2, 1228632926|%e %b %Y, %H:%M %Z|agohover
New post
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License