Which is the smallest square number that, when divided by a cube number, gives a fraction whose simplest form has a numerator (different from 1) that is a cube number and a denominator (different from 1) that is a square number?
The numerator is a square number, which after simplification is a cube number. The general form of such a number is $a^{6} b^{2}$, where $a>1, b \geq 1$.

If $b=1$, then the numerator is $a^{6}$ and the denominator is a cube number that, when simplified by $a^{3}$, results in a square number that is relatively prime to $a$ and different from 1. Therefore, the denominator is $a^{3} c^{6}$, where $c>1$, and $a$ and $c$ are relatively prime numbers. Indeed,

$$
\frac{\left(a^{3}\right)^{2}}{\left(a c^{2}\right)^{3}}=\frac{a^{6}}{a^{3} c^{6}}=\frac{a^{3}}{c^{6}}=\frac{a^{3}}{\left(c^{3}\right)^{2}}
$$

Considering we are looking for the smallest square number, we can ignore the case where $b>1$.

The numerator is the smallest when $a=2$, so the smallest square number sought is 64. (The denominator is the smallest when $c=3$. $\left.\frac{\left(2^{3}\right)^{2}}{\left(2 \cdot 3^{2}\right)^{3}}=\frac{8^{2}}{18^{3}}=\frac{64}{5832}=\frac{8}{729}=\frac{2^{3}}{27^{2}}\right)$.

Géza Biczó (Bp., II., Rákóczi g. II. o. t.)