A number theory problem by Ilham Saiful Fauzi

We need to find all $n$ such that $\displaystyle n-3 | n^3-3$

We know $\displaystyle n-3 | n^3-27$ , So $\displaystyle n-3 | (n^3-3)-(n^3-27) \implies n-3 | 24$

Note we have a total of $16$ divisors of $24$ including negative divisors. The factor 3 helps to determine that the values of $n$ so obtained for each pair of positive and negative divisors is $2\times 3=6$ , For example $n=27$ & $n=-21$ are divisors of 24 which add upto $6$ .

So for $8$ divisors we have sum of all values $\boxed{6\times 8=48}$

Dividing $n^{3} - 3$ by $n-3$ gives a remainder of $\dfrac{24}{n-3}$ . In order for the result to be an integer, the remainder must also be an integer, so all positive and negative factors of $24$ are solutions to $n-3$ .

$24$ has eight positive factors, so there are sixteen solutions to $n-3$ . The sum of all $n-3$ is clearly $0$ ; for every positive solution, its opposite is also a solution.

To find the sum of all $n$ , we just need to add $3$ to all $n-3$ . Because there are sixteen solutions to $n-3$ , the sum of all $n$ is $16\times 3= \fbox{48}$