Solving: $(b+c)^2=2011+bc$ 
Solve $$(b+c)^2=2011+bc$$ for integers $b$ and $c$.

My tiny thoughts:
$(b+c)^2=2011+bc\implies b^2+c^2+bc-2011=0\implies b^2+bc+c^2-2011=0$
Solving in $b$ as Quadratic.$$\implies b=\frac{-c\pm \sqrt{8044-3c^2}} {2}.$$
So $8044-3c^2=k^2$, as $b$ and $c$ are integers. We also have inequalities: $8044>3c^2,8044>3b^2\\ \ \ \ \ 51>c\ \ \ \  , \ \ \ \ 51>b$
How to proceed further. Help.

Consider the ring $R:=\Bbb Z[\rho]\subset\Bbb C$, weher $\rho=\frac{1+i\sqrt3}{2}=e^{2\pi i/6}$ is a sixth root of unity.
Note that $|a+b\rho|^2=(a+b\rho)(a+b\bar \rho)=a^2+ab(\rho+\bar\rho)+b^2\rho\bar\rho=a^2+ab+b^2$. Hence we have a multiplicative map $R\to\Bbb Z$, $z\mapsto|z|^2$ and are looking for all points mapping to $2011$. Note that if $z$ is a solution, then so is $\rho z,\rho^2 z,\ldots \rho^5 z$ as well a their complex conjugates, i.e., solutions come in groups of twelve (unless some of these coincide). Among the twelve solutions of such a group of solutoins, which lie on the vertices of two hexagons in $\Bbb C$, one is in the angle bounded by $[0,\infty)$ and $e^{2\pi i/12}\cdot [0,\infty)$, which just means that $0\le b\le a$ and therefore $\frac 132011\le a^2\le 2011$, i.e., $26\le a\le 44$. For these 19 candidates, we check if $8044-3a^2$ is a square ...
