Diophantine Equation

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 $n$.