jmc

If two numbers give the same remainder when divided by 5, they are said to be equivalent, modulo 5. From $n^2$ to $n^3$, we have multiplied by $n$. Since $n^2$ is equivalent to 4 (modulo 5), and $n^3$ is equivalent to 2 (modulo 5), we are looking for an integer $n$ for which $4\cdot n$ is equivalent to 2, modulo 5. Notice that if $n$ is greater than 4, then we can replace it with its remainder when divided by 5 without changing whether it satisfies the condition. Therefore, we may assume that $0\le n<5$. Trying 0, 1, 2, 3, and 4, we find that only $\boxed{3}$ times 4 leaves a remainder of 2 when divided by 5.