# Pythagorean triples proof

1. Feb 24, 2014

### mikky05v

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?

2. Feb 24, 2014

### Dick

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.

3. Feb 24, 2014

### Dick

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.