Solve in integers the equation
$$
x^{2}+x y+y^{2}=\left(\frac{x+y}{3}+1\right)^{3} \text {. }
$$
We do the trick of setting $a=x+y$ and $b=x-y$. This rewrites the equation as
$$
\frac{1}{4}\left((a+b)^{2}+(a+b)(a-b)+(a-b)^{2}\right)=\left(\frac{a}{3}+1\right)^{3}
$$
where $a, b \in \mathbb{Z}$ have the same parity. This becomes
$$
3 a^{2}+b^{2}=4\left(\frac{a}{3}+1\right)^{3}
$$
which is enough to imply $3 \mid a$, so let $a=3 c$. Miraculously, this becomes
$$
b^{2}=(c-2)^{2}(4 c+1) .
$$

So a solution must have $4 c+1=m^{2}$, with $m$ odd. This gives
$$
x=\frac{1}{8}\left(3\left(m^{2}-1\right) \pm\left(m^{3}-9 m\right)\right) \quad \text { and } \quad y=\frac{1}{8}\left(3\left(m^{2}-1\right) \mp\left(m^{3}-9 m\right)\right) .
$$

For mod 8 reasons, this always generates a valid integer solution, so this is the complete curve of solutions. Actually, putting $m=2 n+1$ gives the much nicer curve
$$
x=n^{3}+3 n^{2}-1 \text { and } y=-n^{3}+3 n+1
$$
and permutations.
For $n=0,1,2,3$ this gives the first few solutions are $(-1,1),(3,3),(19,-1),(53,-17)$, (and permutations).