Question

Find the smallest integer larger than 1 which is a perfect square as well as a perfect cube.

Answer 1

Let us start with any number $n$ and multiply it with itself $6$ times to get a number $N$. We observe that
$N=n\times n\times n\times n\times n\times n=(n\times n)\times (n\times n)\times (n\times n)$
$=\left(n^2\right)\times \left(n^2\right)\times \left(n^2\right)=(n^2{)}^3$.
Thus $N$ is the cube of $n^2$. On the other hand you may also observe that
$N=n\times n\times n\times n\times n\times n=(n\times n\times n)\times (n\times n\times n)$
$=\left(n^3\right)\times \left(n^3\right)=(n^3{)}^2.$
Hence $N$ is also the square of $n^3$. Thus $N$ is both a perfect cube and a perfect square. Taking $n=2$, we get the least number: $N=2\times 2\times 2\times 2\times 2\times 2=64$. We may verify that $64=4^3$ and $64=8^2$.