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Christine
Full Member    Posts: 159 Pythagorean triples   « on: Jun 7th, 2014, 12:20pm » Quote 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: Posts: 156 Re: Pythagorean triples   « Reply #1 on: Jun 8th, 2014, 5:49am » Quote 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
Uberpuzzler      Posts: 1026 Re: Pythagorean triples   « Reply #2 on: Jun 10th, 2014, 8:14pm » Quote 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: Posts: 156 Re: Pythagorean triples   « Reply #3 on: Jun 11th, 2014, 5:03am » Quote 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 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 IP Logged