Cube? A Geometrical Shape?

Let: $a$ and $b$ be positive integers

Let: $c$ be a perfect cube and a perfect square

$c = a^2 = b^3$

Re-arranging this to find $a$ we get:

$a = b^{\frac{3}{2}}$

$a = b^{1 + \frac{1}{2}}$

$a = b \cdot b^{\frac{1}{2}}$

$a = b\sqrt{b}$

Since $a$ is an integer, $b\sqrt{b}$ must also must be an integer. Given that $b$ is an integer and that the square root of a number that isn't a perfect square is always irrational. We can conclude that $b$ must be a square number. The lowest square number above $1$ is $4$ .

$b = 4$

$\therefore a = 4\sqrt{4} = 8$

Solving for $c$ gives us:

$c = a^2 = b^3$

$c = 8^2 = 4^3$

$c = 64 = 64$

$\large \boxed{c = 64}$

Relevant wiki: Perfect Squares, Cubes, and Powers

Let the number be $n$ . We have $n=a^2=b^3$ .Taking square root on both sides we get $a=b^{3/2}$ and taking cube root on both sides we get $b=a^{2/3}$ which won't be integers unless they are sixth powers. So $n=m^6$ . Hence least value of $n$ occurs when $m=2$ i.e. $n=\boxed{64}$ .

In general, if a number is a $n_1$ power, $n_2$ power, $\ldots,n_m$ power, that number will be a $\text{lcm}(n_1,n_2,\ldots,n_m)$ power.

P.S. A trial and error solution.
$2^2 = 4$ but 4 is not a perfect cube. $2^3 = 8$ but 8 is not a perfect square.
$3^2 = 9$ but 9 is not a perfect cube. $3^3 = 27$ but 27 is not a perfect square.
$4^2 = 16$ but 4 is not a perfect cube. $4^3 = 64 = 8^2$ . So, the answer is $\color{#69047E}{\boxed{64}}$ .

$n = (2\cdot 2\cdot 2)\cdot (2\cdot 2\cdot 2)= (2\cdot 2)\cdot (2\cdot 2)\cdot (2\cdot 2)$ = 64.

In general, the number of 2's needed is the least common multiple of the powers. For example, if the question were $n=a^4=b^6$ , the lcm is 12 and so $n=(2\cdot 2\cdot 2)^4 = (2\cdot 2)^6$ .

While the factorization works with any other base besides 2, it would give a larger result, and the question asks for the smallest n.