Do Pythagorean Triples Generate All Possible Solutions?

[mikky05v]

Homework Statement

prove the following proposition:
Let P be the set of pythagorean triples; that is,
P= {(a,b,c) : a, b, c \in Z and a^{2}+b^{2}=c^{2}}
and let T be the set
T= {(p,q,r) : p=x^{2}-y^{2}, q= 2xy, and r = x^{2}+y^{2} where x,y\in Z}

show also that T ≠ P … that is that T is a “proper subset” of P … that is that there’s at least 1 member of P that is NOT in T.

Homework Equations

The Attempt at a Solution

I understand what it's asking but i have no idea how to begin proving it. I thought at first to try and let x\in P and go from there but then I didn't know how to work with that idea at all. Can someone point me the right direction on figuring this out?

 

[Dick]

Homework Statement

prove the following proposition:
Let P be the set of pythagorean triples; that is,
P= {(a,b,c) : a, b, c \in Z and a^{2}+b^{2}=c^{2}}
and let T be the set
T= {(p,q,r) : p=x^{2}-y^{2}, q= 2xy, and r = x^{2}+y^{2} where x,y\in Z}

show also that T ≠ P … that is that T is a “proper subset” of P … that is that there’s at least 1 member of P that is NOT in T.

Homework Equations

The Attempt at a Solution

I understand what it's asking but i have no idea how to begin proving it. I thought at first to try and let x\in P and go from there but then I didn't know how to work with that idea at all. Can someone point me the right direction on figuring this out?

Use algebra to show your formula generates Pythagorean triples, i.e. show p^2+q^2=r^2. That's easy enough. To show it doesn't generate all of them takes a little more guesswork. You have to find a triple is doesn't generate. Here's a hint. Multiply some of the known triples by a constant.

 

[Dick]

Homework Statement

prove the following proposition:
Let P be the set of pythagorean triples; that is,
P= {(a,b,c) : a, b, c \in Z and a^{2}+b^{2}=c^{2}}
and let T be the set
T= {(p,q,r) : p=x^{2}-y^{2}, q= 2xy, and r = x^{2}+y^{2} where x,y\in Z}

show also that T ≠ P … that is that T is a “proper subset” of P … that is that there’s at least 1 member of P that is NOT in T.

Homework Equations

The Attempt at a Solution

I understand what it's asking but i have no idea how to begin proving it. I thought at first to try and let x\in P and go from there but then I didn't know how to work with that idea at all. Can someone point me the right direction on figuring this out?

Use algebra to show your formula generates Pythagorean triples, i.e. show p^2+q^2=r^2. That should be easy enough. To show it doesn't generate all of them takes a little more guesswork. You have to find a triple is doesn't generate. Here's a hint. Multiply some of the known triples by a constant.