# Onto Functions and Integers

The function $f: ℝ → ℝ$ defined by $f(x) = x^{3}$ is onto because for any real number $r$, we have that $\sqrt[3]r$ is a real number and $f(\sqrt[3]r)=r$. Consider the same function defined on the integers $g: ℤ → ℤ$ by $g(n) = n^3.$ Explain why $g$ is not onto $ℤ$ and give one integer that $g$ cannot output.

I can't think of any integer that cannot be cubed, so this problem has me confused.

The problem is not that we can't cube some integer, its that not every integer has an integer cube root. Consider $5\in\mathbb{Z}.$ To say that $5$ is in the image of $g$ is to say that $5^{1/3}\in \mathbb{Z}.$ As we know, $5^{1/3}\approx1.71\not\in \mathbb{Z}$.
You don't have to think of an integer that cannot be cubed, but of an integer from which you cannot extract a cubic root (staying in $\mathbb{Z}$, of course, it will have a cubic root in $\mathbb{R}$).
For instance, $g(x)=2$ doesn't have any solution if you ask that $x\in \mathbb{Z}$.