$\textbf{Question:}$Find all pairs of positive integers $a,b$ such that $\frac{a^3+b^3}{ab+4}=2020$
First of all, I factored $a^3+b^3$ as $(a+b)(a^2-ab+b^2)$.Since $101$ divies $2020$ and if we take it away what remains is $20$ which is a pretty small number.So,I thought of considering two cases $ 101 \mid a+b$ or $101 \mid a^2-ab+b^2$.Doing that,what remains in the $L.H.S$ is bounded by $20$.So,I thought that can be taken care of using some case-work .But there seems to be too many cases.
Any kind of hint or solutions are appreciated.Thanks in advance.