wu :: forums
wu :: forums - Pythagorean triples

Welcome, Guest. Please Login or Register.
Nov 28th, 2023, 6:41pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   medium
(Moderators: Grimbal, towr, Eigenray, Icarus, SMQ, william wu, ThudnBlunder)
   Pythagorean triples
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Pythagorean triples  (Read 1426 times)
Christine
Full Member
***





   


Posts: 159
Pythagorean triples  
« on: Jun 7th, 2014, 12:20pm »
Quote Quote Modify Modify

Pythagorean triples (a, b, c)
then a^2 + b^2 = c^2
 
How to prove that
c^2 - a*b and c^2 + a*b can both be expressed as a sum of two squares?
IP Logged
0.999...
Full Member
***





   


Gender: male
Posts: 156
Re: Pythagorean triples  
« Reply #1 on: Jun 8th, 2014, 5:49am »
Quote Quote Modify Modify

c^2-ab=( 1/2(c+(b-a)) )^2 + ( 1/2(c-(b-a)) )^2 and c^2+ab=( 1/2(c+(b+a)) )^2 + ( 1/2(c-(b+a)) )^2
 
I really only examined two cases and extrapolated. Perhaps someone else will be able to offer more insight.
IP Logged
rloginunix
Uberpuzzler
*****





   


Posts: 1026
Re: Pythagorean triples  
« Reply #2 on: Jun 10th, 2014, 8:14pm »
Quote Quote Modify Modify

The search for all the positive integer solutions of the algebraic equation x^2 + y^2 = z^2 has the geometric equivalent of finding all the right triangles with the integral side lengths. If I remember correctly early on the Babylonians raised that question which also interested the Greek geometers of the time. Pythagoras himself is credited with the following solution:
 
a = 2n + 1, b = 2n^2 + 2n, c = 2n^2 + 2n + 1
 
for his a^2 + b^2 = c^2 triple (n is an arbitrary positive integer).
 
My idea leads to lots of writing but is simple:
 
a*b = (2n + 1)(2n^2 + 2n) = 4n^3 + 4n^2 + 2n^2 + 2n = 4n^3 + 6n^2 + 2n
 
c^2 = (2n^2 + 2n + 1)^2 = 4n^4 + 4n^2(2n + 1) + 4n^2 + 4n + 1 = 4n^4 + 8n^3 + 8n^2 + 4n + 1
 
Rearrangement of your difference, for example:
 
c^2 - a*b =  4n^4 + 8n^3 + 8n^2 + 4n + 1 - 4n^3 - 6n^2 - 2n = 4n^4 + 4n^3 + 2n^2 + 2n + 1 = 4n^4 + 4n^3 + n^2 + (n^2 + 2n + 1) =  
 
 =  4n^4 + 4n^3 + n^2  + (n + 1)^2 = n^2(4n^2 + 4n + 1) + (n + 1)^2 = n^2(2n + 1)^2 + (n + 1)^2 =
 
 = (n(2n + 1))^2 + (n + 1)^2
 
You can rename the terms "n(2n + 1)" and "n + 1" as some other integers of your liking and you have your sum of two squares.
 
Perhaps now you can try your hand at "c^2 + a*b".
IP Logged
0.999...
Full Member
***





   


Gender: male
Posts: 156
Re: Pythagorean triples  
« Reply #3 on: Jun 11th, 2014, 5:03am »
Quote Quote Modify Modify

rloginlinux, while it is an infinite class of solutions, the parametrization does not cover all primitive solutions. An example of one that is missed is (8,15,17).
 
In the textbook on Number Theory by Ireland and Rosen, excercise 1.23, one shows that the following two-variable paramatrization, covers all primitive pythagorean triples (a,b,c) up to a permutation of a and b:
a = 2uv, b = v^2-u^2, c = v^2+u^2.
 
Using this parametrization, the calculations are rather nice:
c^2-ab=v^4+u^4+2u^2v^2-2uv^3+2u^3v=v^4-2uv^3+u^2v^2+u^4+2u^3v+u^2v^2=v^2(v-u)^2+u^2(v+u)^2. The calculation in the case c^2+ab is the same, except with u and v reversed.
IP Logged
Immanuel_Bonfils
Junior Member
**





   


Posts: 114
Re: Pythagorean triples  
« Reply #4 on: Jul 30th, 2014, 8:14pm »
Quote Quote Modify Modify

on Jun 8th, 2014, 5:49am, 0.999... wrote:
c^2-ab=( 1/2(c+(b-a)) )^2 + ( 1/2(c-(b-a)) )^2 and c^2+ab=( 1/2(c+(b+a)) )^2 + ( 1/2(c-(b+a)) )^2
 

Are you sure about this?
IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright 2000-2004 Yet another Bulletin Board