Find all integers n (positive, negative, or zero) so that $n^3 – 1$ is divisible by $n + 1$

Question

I've tried to rearrange $n^3 - 1$ (turning it into $(n-1)(n^2+n+1)$) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.

Answer 1

John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:

$$n^3-1=(n+1)(n^2-n+1)-2$$

Dividing by $n+1$ gives $$\frac{n^3-1}{n+1}=(n^2-n+1)-\frac{2}{n+1}$$

Since the left side & the first term on the right are integers, then $\frac{2}{n+1}$ must also be an integer. So, $\frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$

Answer 2

If $n + 1$ is a factor, then note that $n \equiv -1 \pmod{n + 1}$. As such, $n^3 - 1 \equiv \left(-1\right)^3 - 1 = -2 \pmod{n + 1}$. This can only be the case if $n + 1 = \pm 2$ or $n + 1 = \pm 1$, i.e., $n = 0, 1, -2, -3$.