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 $2020$
 A: 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...  
A: Note that if $p\mid a^2-ab+b^2$, where $p$ is a prime natural number s.t. $p\equiv 2\pmod{3}$, then $p\mid a$ and $p\mid b$.  For $p=2$, the claim is easily seen by inspection.  Let now $p>2$.  We prove by contradiction.  Suppose that $p\nmid a$ or $p\nmid b$.  It follows immediately that $p\nmid a$ and $p\nmid b$. Since
$$4(a^2-ab+b^2)=(2a-b)^2+3b^2\equiv0 \pmod{p},$$
we have $x^2\equiv-3\pmod{p}$, where $x=(2a-b)c$ if $c$ is an inverse of $b$ modulo $p$.  Consequently,
$$\left(\frac{-3}{p}\right)=1.$$
By quadratic reciprocity,
$$\left(\frac{p}{3}\right)=\left(\frac{p}{-3}\right)=1.$$
Hence $p\equiv 1\pmod{3}$, which is a contradiction.
From $\frac{a^3+b^3}{ab+4}=2020$, we get
$$2020(ab+4)=a^3+b^3=(a+b)(a^2-ab+b^2).$$
If $101\mid a^2-ab+b^2$, then $101\mid a$ and $101\mid b$ by the paragraph above.  Thus $101^3\mid (a+b)(a^2-ab+b^2)$, but clearly $101^3\nmid 2020(ab+4)$.  Hence, $101\mid a+b$.
Similarly, $5\mid a+b$.
If $2\mid a^2-ab+b^2$, then $2\mid a$ and $2\mid b$.  Write $a=2u$ and $b=2v$, then
$$1010(uv+1)=(u+v)(u^2-uv+v^2).$$
If $2\mid u^2-uv+v^2$, then $2\mid u$ and $2\mid v$.  Therefore $8\mid (u+v)(u^2-uv+v^2)$, but clearly $8\nmid 1010(uv+1)$.  Hence $2\mid u+v$.  Consequently $4\mid a+b$.  On the other hand, if $2\nmid a^2-ab+b^2$, then $4\mid a+b$ as $4\mid 2020$.  In any case, $4\mid a+b$.  This means $$a+b=4\cdot 5\cdot 101 \cdot k=2020 k$$ for some positive integer $k$.
Hence
$$k=\frac{a+b}{2020}=\frac{ab+4}{a^2-ab+b^2}.$$
Clearly $a\ne b$.  Since $a+b=2020k$, $a\equiv b\pmod{2}$ so $(a-b)^2\ge 4$.  If $(a-b)^2>4$, then $a^2-2ab+b^2>4$, making $ab+4<a^2-ab+b^2$, so
$$k=\frac{ab+4}{a^2-ab+b^2}<1,$$
which is a contradiction.  Hence, $(a-b)^2=4$, so that $ab+4=a^2-ab+b^2$, making $k=1$.  Thus we have $a+b=2020k=2020$ and $a-b=\pm\sqrt4=\pm2$.  This gives 
$$(a,b)=(1009,1011)\vee (a,b)=(1011,1009),$$
and both are solutions.  In fact, these two solutions are the only integer solutions (positive or negative) to the required equation.
