Which is the six-digit number that is the square of a non-perfect power and when divided by a cube number, gives a fraction less than one, where in its simplest form, the numerator is the cube of a non-perfect power, and the denominator is the square of a non-perfect power. (The number one should be considered a perfect power.)
Let the sought numerator be denoted by $x^{2}$. Since $x$ is not a perfect power, there is a prime factor among its prime factors whose exponent is not divisible by 3. Denoting the product of such prime factors by $b$, $x$ can be written in the form $x=a^{3} b$, where $a$ and $b$ are relatively prime numbers and $b \neq 1$.

However, $a \neq 1$ also holds, because after simplification, a perfect cube must be obtained, which is impossible if $a=1$. Therefore, only numerators of the form $a^{6} b^{2}$, where $a$ and $b$ are relatively prime, $b$ is not a perfect cube, and $a \neq 1, b \neq 1$, satisfy our conditions.

Such a form of numerator always works.

Because if $a$ is not a perfect power, it is sufficient to choose the denominator as the product $a^{3} b^{6} c^{6}$, where $b$ and $c$, as well as $a$ and $c$, are relatively prime, $c$ is not a perfect square, but is large enough so that the value of the fraction is less than 1.

Indeed,

$$
\frac{a^{6} b^{2}}{a^{3} b^{6} c^{6}}=\frac{\left(a^{3} b\right)^{2}}{\left(a b^{2} c^{2}\right)^{3}}=\frac{a^{3}}{b^{4} c^{6}}=\frac{a^{3}}{\left(b^{2} c^{3}\right)^{2}}
$$

If, however, $a$ is a perfect power, write $a$ in the form $m \cdot n$ (where possibly $m=n$) and choose the denominator as the product $m^{3} n^{6} b^{6} c^{6}$. In this case,

$$
\frac{m^{6} n^{6} b^{2}}{m^{3} n^{6} b^{6} c^{6}}=\frac{\left(m^{3} n^{3} b\right)^{2}}{\left(m n^{2} b^{2} c^{2}\right)^{3}}=\frac{m^{3}}{b^{4} c^{6}}=\frac{m^{3}}{\left(b^{2} c^{3}\right)^{2}}
$$

Therefore, the solutions to our problem are all 6-digit numbers that are of the form

$$
a^{6} b^{2}
$$

where $a$ and $b$ are relatively prime, $a \neq 1, b \neq 1$, and $b$ is not a perfect cube.

Since according to the condition

$$
10^{5} < a^{6} b^{2} < 10^{6}
$$

we need to find all such pairs $(a, b)$. For $a \leq 7$, we can find suitable $b$ values, but for $a > 7$ we do not find solutions. Therefore, the total number of solutions is: 70.