... +bi) 2 = (a 2 −b 2 )+2abi. Comparingreal and imaginary parts yields x = a 2 − b 2 ,y = 2ab, and since x 2 + y 2 = z 2 ,z =a 2 +b 2 . Conversely, it is easily shown that for any a,b ∈ Z, (a 2 −b 2 , ... Z, (a 2 −b 2 , 2ab,a 2 +b 2 )is a solution of x 2 + y 2 = z 2 . We thus get all Pythagorean triples. It is easy to singleout the primitive ones among them.Coming back to x 2 + 2 = y3, we ... sideand get (x +√2i)(x −√2i) = y3, an equation in the domain D ={a + b√2i :a, b ∈ Z}. Here too we can show that (x +√2i, x −√2i) = 1, hence x +√2i andx −√2i are cubes in D....