Solve the following equation in integers:

$3{\mathrm{x}}^3-{\mathrm{x}}^2\mathrm{y}-{\mathrm{xy}}^2+3{\mathrm{y}}^3=2013$

Then the value of ${\mathrm{x}}^2+{\mathrm{y}}^2$ is____.

Observe that setting $\mathrm{x}=-\mathrm{y}$ makes the left hand side 0 . Hence $(\mathrm{x}+\mathrm{y})$ is a factor of the left hand side. We find that the equation is equivalent to

$(\mathrm{x}+\mathrm{y})\left(3{\mathrm{x}}^2-4\mathrm{xy}+3{\mathrm{y}}^2\right)=(\mathrm{x}+\mathrm{y})\left(3(\mathrm{x}+\mathrm{y}{)}^2-10\mathrm{xy}\right)=2013.$

From here, the naïve way to proceed is to simply note that the problem is now reduced to checking cases, as the prime factorization of 2013 is $3\cdot 11\cdot 61.$.

This implies that $(\mathrm{x}+\mathrm{y})\le 4026<4096={16}^3$. Hence $\mathrm{x}+\mathrm{y}<16$. Since $\mathrm{x}+\mathrm{y}$ is a positive factor of 2013 , we only have three possibilities for $\mathrm{x}+\mathrm{y}$, specifically 1,3 , or 11 . Then computing

$(\mathrm{x}-\mathrm{y}{)}^2=\frac{2·2013-(\mathrm{x}+\mathrm{y}{)}^3}{5(\mathrm{x}+\mathrm{y})}$

gives $(\mathrm{x}-\mathrm{y}{)}^2=805\mathrm{y}$ when $\mathrm{x}+\mathrm{y}=1,(\mathrm{x}-\mathrm{y}{)}^2=\frac{1333}{5}$ when $\mathrm{x}+\mathrm{y}=3$ and $(\mathrm{x}-\mathrm{y}{)}^2=49$, when $\mathrm{x}+\mathrm{y}=11$. The only case in which $\mathrm{x}-\mathrm{y}$ is an integer is the case when $\mathrm{x}+\mathrm{y}=11$, giving $\mathrm{x}-\mathrm{y}=\pm 7$. Solving the linear systems that result, we obtain the solutions $(\mathrm{x},\mathrm{y})=(2,9)$ or $(9,2).$