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Consider two cases, z0 even and z0 odd. If z0 is even, then x0 and y0 are odd. But then x0 + y0 and x0 − y0 are both even. Let 2p = x0 + y0 and 2q = x0 − y0. Then x0 =p+q and y0 = p-q. It follows from gcd(x0, y0) = 1 that gcd(p, q) = 1. Parity arguments show that p not≡ q (mod 2). Lastly, z0^3 = 2p(p^2 + 3q^2). Summarizing, there exist p and q so that: (a) p and q are positive (b) gcd(p, q) = 1 (c) p not≡ q (mod 2) (d) 2p(p^2 + 3q^2) is a cube Answer the following: 1. Use gcd(x0, y0) = 1 to prove that gcd(p, q) = 1. 2. Prove that p not≡ q (mod 2). 3. Prove that z0^3 = 2p(p^2 + 3q^2). Note x0 = x not, y0 = y not, z0 = z not
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