Are P and Q equal for Pythagorean triples and parametric 3-tuples?

[honestrosewater]

Sorry if this doesn't come out right- figuring out how to state it is part of my problem.
Let P be the set of all 3-tuples (a, b, c) in N such that a² + b² = c² and a < b.
Let Q be the set of all 3-tuples (x, y, z) in N such that x = m(2n + 1), y = m(2n² + 2n), and z = m(2n² + 2n + 1) for some m and n in N.
I want to figure out if P = Q. I know that for all sets S and T, (S = T) is equivalent to (S is a subset of T and T is a subset of S), but I don't know where to begin. Should I try to prove that for all 3-tuples t in N, if t is in P, then t is in Q and if t is in Q, then t is in P? Should I state it a different way (I tried, but it seemed more complicated)?

 

[snoble]

What you want to do is take an arbitrary element of one set and show it belongs in the other set. One way appears easy. Take a 3-tuple in Q (x,y,z). Leaving m and n undefined consider $$x^2 + y^2$$. Does it equal $$z^2$$. The other way is harder. Take a 3-tuple in P (a,b,c). You know that $$a^2+b^2= c^2$$. Can you find m and n integers such that the first two requisites are satisfied? If you can the 3rd requisite is satisfied automatically by the work done earlier.

Also numbers that satisfy $$a^2+b^2 =c^2$$ have been completely classified so looking that up may help.

Good luck,
Steven

 

[Hurkyl]

I want to figure out if P = Q. I know that for all sets S and T, (S = T) is equivalent to (S is a subset of T and T is a subset of S), but I don't know where to begin. Should I try to prove that for all 3-tuples t in N, if t is in P, then t is in Q and if t is in Q, then t is in P? Should I state it a different way (I tried, but it seemed more complicated)?

Yes, what you have stated sounds correct, and is a typical approach. One alternative approach is:

x is in P ==> x is in Q
x isn't in P ==> x isn't in Q

 

[honestrosewater]

Yeah, Q is a subset of P was easy- just manipulation. I couldn't get anywhere trying to prove the other half, so I gave up, went searching, and found (8, 15, 17). Eh. x = d(n² - m²), y = d2nm, and z = d(n² + m²), for d in N and relatively prime n > m in N supposedly works (I haven't checked), but by this my counterexample above is actually (15, 8, 17), so I guess I just need to lose a < b - my brain is pudding now, so I'll look at it later. Thanks for the help.