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.
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.
Where are r,s,t,u coming from. Do you mean given a solution to
$\displaystyle x^2+y^2=t^2$
and
$\displaystyle t^2+z^2=w^2$
Find another solution, r,s,u,v,p from this solution such that:
$\displaystyle r^2+s^2=u^2$
and
$\displaystyle u^2+v^2=p^2$
with $\displaystyle (r,s,u)=(u,v,p)=1$
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?
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.
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 $\displaystyle m,n\in \mathbb{N}$ with $\displaystyle m<n$.
Then $\displaystyle (m^2-n^2,2mn,m^2+n^2)$ 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.