Question P.1: Find all solutions in integers of x³ + 2y³ = 4z³....

Solution 1. First note that if (x, y, z) is a solution in integers, then x must be even, say x = 2w. Substituting this, dividing by 2, and subtracting y³, we see that (−y)³ + 2z³ = 4w³ , so (−y, z, w) = (−y, z, x/2) is another solution. Now repeat this process to get (x, y, z), (−y, z, x/2), (−z, x/2, −y/2), (−x/2, −y/2, −z/2) as successive integer solutions. Conclusion: If (x, y, z) is a solution, then so is (−x/2, −y/2, −z/2), and in particular, x, y, and z are all even. But if x, y, and z were not all zero, we could keep replacing (x, y, z) by (−x/2, −y/2, −z/2) and eventually arrive at a solution containing an odd integer, a contradiction.

Solution 2. Suppose (x, y, z) is a nonzero solution for which |x|³ + 2|y|³+4|z|³ is minimized. Clearly x is even, say x = 2w. We have 8w³ + 2y³ = 4z³ or (−y)³ + 2z³ = 4w³ . Then (−y, z, w) is a nonzero solution and
| − y|^3 + 2|z|^3 + 4|w|^3 =\frac{|x|^3 + 2|y|^3 + 4|z|^3}{2},
a contradiction. Therefore (0, 0, 0) is the only solution.