Find the all positive integer solutions $(a,b)$ to $\frac{a^3+b^3}{ab+4}=2020$. 
Find the all positive integer solutions of given equation
  $$\frac{a^3+b^3}{ab+4}=2020.$$

I find two possible solutions, namely $(1011,1009)$ and $(1009,1011)$, but the way I solve  the equation was messy and I don't know if there are any other solutions.
Source: Turkey $1.$ TST for IMO $2020$

Write for ease $n=2020$ and let $c=a+b$. As $b=c-a$ we get a following quadrtatic equation on $a$:  $$(3c+n)a^2-(3c+nc)a+c^3-4n=0$$
So it discriminat must be a perfect square $d^2$ (as it has solution in $\mathbb{Z}$):
$$d^2 = -3c^4+2nc^3+n^2c^2+48nc+16n^2\;\;\;\;\;(*)$$ from here we get $$\boxed{2n\mid d^2+3c^4}$$
Now what can we say about $c$?


*

*If $5\nmid c$ then $c^4\equiv_5 1$ so $d^2+3\equiv _5 0$ which is not possible. So $5\mid c$. 

*Since $8\mid d^2+3c^2$, $d$ and $c$ nust be the same parity. Say both are odd. Since for each odd $x$ we have $x^2\equiv_8 1$ we get $$ 0\equiv _8 d^2+3c^4 \equiv_8 1+3$$ A contradiction. So $c$ and $d$ are even. Since $8\mid 3c^4$ we have $8\mid d^2$ so $4\mid d$. 

*If $101\nmid c$ then $$d^2c^{-4} \equiv_{101} -3\implies \Big({-3\over 101}\Big)=1$$
But $$\Big({-3\over 101}\Big) = \Big({-1\over 101}\Big)\Big({3\over 101}\Big) = 1\cdot \Big({101\over 3}\Big)(-1)^{{3-1\over 2}{101-1\over 2}} = -1$$
A contradiction again, so $101\mid c$
So $$\boxed{1010\mid c}$$
Now suppose $c>n$. From $(*)$ we get:  \begin{align}3c^4&\leq 2nc^3+n^2c^2+48nc+16n^2\\
&< 2(c-1)c^3+(c-1)c^2+64c^2\\
& = 3c^4-4c^4+65c^2
\end{align}
and now we have $4c^3<65c^2$, a contradiction. So $c\leq 2020$. 
So $c\in\{1010,2020\}$ and we check both values manualy...  
