It started with the following Diophantine equation:
Find all positive integers $x, y$ such that: $$2(x^3 - x) = y^3 - y $$
First we see that $(x,y) = (1,1)$ is a solution.
Consider $x,y \ge2$, we have: $$y^3-y = (\sqrt[3]{2}x)^3 - 2x <(\sqrt[3]{2}x)^3 - \sqrt[3]{2}x$$ Because $f(x) = x^3-x$ is strictly increasing on $\left[1;+\infty \right)$, we have $y<\sqrt[3]{2}x$. Suppose $y \le \sqrt[3]{2}x-1$. Then: $$2x^3 - 2x = y^3 - y \le (\sqrt[3]{2}x-1)^3 - (\sqrt[3]{2}x - 1) = 2x^3 -3\sqrt[3]{4}x^2 +2\sqrt[3]{2}x \iff 3\sqrt[3]{4}x^2 \le (2+2\sqrt[3]{2})x $$
Which is false for $x\ge 2$, so $\sqrt[3]{2}x >y > \sqrt[3]{2}x-1$. Since $y$ is an integer, we must have $y = \lfloor \sqrt[3]{2}x \rfloor$. I then plugged in some values of $x$ and got $(x,y) = (4,5)$ and it seemed like $\lfloor \sqrt[3]{2}x \rfloor ^3 - \lfloor \sqrt[3]{2}x \rfloor < 2(x^3-x)$ for $x >4$ but I have no idea whether it's true or not.
How can we prove or disprove the above inequality? (Also, is there a simpler approach to the Diophatine equation?)