Find all $a\in\mathbb{N}$ such that $3a+6$ divides $a^2+11$

Question

Find all $a\in\mathbb{N}$ such that $3a+6$ divides $a^2+11$

This problem has stumped me. I don't even know where to begin solving it. I know the solutions will be all $a$ such that $$\frac{a^2+11}{3a+6}=k$$ with $k\in\mathbb{Z}$

But I really don't know how to follow from here

Answer 1

If $3a+6$ divides $a^2+11$, then $3a+6$ divides $3a^2+33$.

$3a+6$ also divides $3a^2+6a$, so this means $3a+6$ divides $6a-33$, or $a+2$ divides $2a-11$.

$a+2$ also divides $2a+4$, so this means $a+2$ divides $15$.

$15$ does not have many factors, so I leave you to check them.