Problem:
$ x^2 + (y - \sqrt[3]{x^2}) = 1 $
Solution:
We have $ x^2 + y - 1 = \sqrt[3]{x^2} $, therefore $ (x^2 + y - 1)^3 = x^2 $, therefore $ x^2 $ must also be a cube.
The smallest non-trivial number that is both a square and a cube is 64, in that case, we have $ x = 8 $ and $ y = -59 $. In general, we can have $ x = n^3 $ and $ y = n^2 - n^6 + 1 $ for all integer $ n $.
$ x^2 + (y - \sqrt[3]{x^2}) = 1 $
Solution:
We have $ x^2 + y - 1 = \sqrt[3]{x^2} $, therefore $ (x^2 + y - 1)^3 = x^2 $, therefore $ x^2 $ must also be a cube.
The smallest non-trivial number that is both a square and a cube is 64, in that case, we have $ x = 8 $ and $ y = -59 $. In general, we can have $ x = n^3 $ and $ y = n^2 - n^6 + 1 $ for all integer $ n $.
