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.

With what you have done so far we can just try the possibilities.  We can ask that $c \ge 0$ and recognize that we can change the signs of both $b$ and $c$ from a solution to get another.  $b$ and $c$ can also be interchanged.  I find $(10,39)$ and $(-10,49)$ as base solutions.  A spreadsheet with copy down makes it simple.
