# Pythagorean Triplets

• Jul 12th 2009, 10:33 AM
diddledabble
Pythagorean Triplets
Given x^2 + y^2 = t^2 and t^2 + z^2 = w^2
where (x,y,t)=(t,z,w)=1 and t,x,y,z,w>0
Show that given one solution to the equation a person can always select r,s,u, and v from this solution that will result in a new solution which has (x,y,t)=(t,z,w)=1.
• Jul 22nd 2009, 10:04 AM
diddledabble
Anyone out there able to help at all with Pythagorean Triples? I really need to understand these and my textbook doesn't help at all. Make up your own example to explain if it helps.
• Jul 22nd 2009, 11:36 AM
lancekam
Where are r,s,t,u coming from. Do you mean given a solution to
$x^2+y^2=t^2$
and
$t^2+z^2=w^2$

Find another solution, r,s,u,v,p from this solution such that:
$r^2+s^2=u^2$
and
$u^2+v^2=p^2$
with $(r,s,u)=(u,v,p)=1$
• Jul 22nd 2009, 11:59 AM
diddledabble
I think you have it right lancekam. But that is all my book tells me. There is nothing else. The only info in the chapter is the proof of pythagorean triples. I am not sure how they expect you to read it and then apply it to other stuff that isn't immediately similar. Do you have any ideas?
• Jul 31st 2009, 11:12 AM
alexmahone
Quote:

Originally Posted by lancekam
Where are r,s,t,u coming from. Do you mean given a solution to
$x^2+y^2=t^2$
and
$t^2+z^2=w^2$

Find another solution, r,s,u,v,p from this solution such that:
$r^2+s^2=u^2$
and
$u^2+v^2=p^2$
with $(r,s,u)=(u,v,p)=1$

An interesting identity:
.
$(x^2+y^2)(t^2+z^2) \;=\;\begin{Bmatrix}(xt-yz)^2 + (xz + yt)^2 \\ (xt+yz)^2 + (xz-yt)^2 \end{Bmatrix}$

Does this help?
• Jul 31st 2009, 12:19 PM
diddledabble
Not really because I don't know where to go or what to do with it.
• Jul 31st 2009, 03:42 PM
alunw
This made no sense to me initially. What are r,s,u,v? Are they supposed to be the numbers that you plug into the usual formual for getting Pythagorean triples and are we supposed to pick r,s,u,v from four of x,y,z,t,w. If so you should say so.
It is easy to see that if the triples (x,y,t) and (t,z,w) are both proper Pythagorean triples then we have t=a^2+b^2 from some (a,b)=1 and also t=c^2-d^2 for some (c,d)=1 Surely t=1 mod 4 which means that c is odd and d is even

Let's see.
3,4,5 comes from (2,1) with 3=2^2-1^2,4=2*2*a,5=2^2+1^1
5,12,13 comes from (3,2) with 5=3^2-2^2,12=2*2*3,13=3^2+2^2

(4,3) gives us the triple (7,24,25)
(13,12) gives us the triple (25,312,313)

(24,7) gives us the triple 527,336,625
(313,312) gives us some triple with 625.

OK this starts to seem plausible.
Now we had t = (a^2+b^2) = (c^2-d^2)
We are going to pick 2*a*b and a^2-b^2 as the generators of the first of the new triples, and 2*c*d and c^2+d^2 as the generators of the second triple
First we need to show that (2*a*b)^2 + (a^2-b^2)^2 = (c^2+d^2)^2-(2*c*d)^2
LHS is 4*a^2*b^2+a^4-2*a^2*b^2+b^4 = (a^2+b^2)^2
RHS is c^4+2*c^2*d^2+d^4-4*c^2*d^2 = (c^2-d^2)^2
So clearly we have LHS=RHS as both are t^2
Also since both pairs of numbers we picked meet the condition to generate triples, (i.e. one odd and one even number with hcf=1) then we have a new pair of triples of the required form.
• Aug 5th 2009, 08:58 AM
diddledabble
Any other explanation on triples. For example if $x^2+y^2=130$ How would you determine that x=7 and y=9?
• Aug 5th 2009, 11:54 AM
Gamma
Well then I suggest you do my suggestion above and consider reduction mod 10 is probably the easiest. Also it may be of use to you to know that every Pythagorean Triple is generated as follows.

Let $m,n\in \mathbb{N}$ with $m.
Then $(m^2-n^2,2mn,m^2+n^2)$ is a Pythagorean Triple. Of course this method takes a while and yields lots of non primitive ones. I suggest memorizing the squares of integers up to 20 or 25 before the exam, it could save you a lot of time. Take your c^2 and subtract some squares from it and see if you recognize it as a perfect square, that is how I did it when you sent me the question the first time.