The wooden cube

The surface of a square appearing on the cube is $x^2$ and the volume of the cube is $x^3$. They are all integers by assumption. We will show that $x$ is indeed an integer.

Since $\displaystyle x=\frac{x^3}{x^2}$, it must be rational. Let $p$ be a prime number and $v_p(y)$ the $p$-adic valuation of a rational number $y$. Then one gets that $v_p(x)$ is well-defined and that $v_p(x^2)=2v_p(x)$ is an even positive integer because $x^2$ is a positive integer. Hence, for all prime $p$, $v_p(x)$ is a positive integer. Hence $x$ is an integer.