Exercise 11. Determine all integers \( a \) such that \( a-3 \) divides \( a^3-3 \).
Solution to Exercise 11

We know that $a-3 \mid a^{3}-3^{3}$, so it divides the difference

$$
a-3 \mid a^{3}-3-\left(a^{3}-3^{3}\right)=27-3=24
$$

Conversely, it is sufficient that $a-3$ divides 24 for it to divide $24+a^{3}-3^{3}=a^{3}-3$. The solutions are therefore exactly the divisors of 24 to which we add 3. We have therefore

$$
a-3 \in\{-24,-12,-8,-6,-4,-3,-2,-1,0,1,2,3,4,6,8,12,24\}
$$

and $a \in\{-21,-9,-5,-3,-1,0,1,2,3,4,5,6,9,11,15,27\}$.

## Comments from the graders:

The exercise was generally well solved; the most common mistake is forgetting some divisors of 24.