<s>[INST] 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.
 [/INST] Decided to see what I could do in my head.
$2011 \equiv 1 \pmod 3$ is a prime. This means it has both expressions as $u^2 + uv + v^2$ and $x^2 + 3 y^2.$ Note that $45^2 = (4 \cdot 5) \cdot 100 + 25 = 2025$ is too large for $x^2.$ Then $44^2 = 45^4 - 2 \cdot 45 + 1 = 2025 - 90 + 1 = 1935 + 1 = 1936$ is small enough. Then $2011 - 1936 = 11 + 64 = 75 = 3 \cdot 25 = 3 \cdot 5^2.$
So far we have $$ 2011 = 44^2 + 3 \cdot 5^2  $$
The order I can remember is $r^2 + rs + k s^2$ when $r = p - q, s = 2q$ gives  $p^2 - 2pq + q^2 + 2 p q - 2 q^2 + 4kq^2 = p^2 + (4k-1)q^2. $ This time $k=1.$   That is, $p^2 + 3 q^2 = r^2 + rs + r^2$ when $r = p-q, s = 2 q.$ we got $p = 44, q = 5.$ So $r = 44 - 5, s = 2 \cdot 5,$ or 
$$ r^2 + rs +  s^2 = 2011, \; \; r = 39,  \; \; s = 10. $$ 
It happens that $x^2 + xy + y^2$ and $u^2 + 3 v^2$ integrally represent exactly the same numbers. This is the only time this happens with positive forms. The other time is $x^2 + xy - y^2$ and $u^2 -5 v^2.$
As soon as we get to $x^2 + xy + k y^2$ with $|k| > 1$ and $1-4k$ not a square, there is a difference. For example, $x^2 + xy + 2 y^2$ represents more numbers (namely those $2 \pmod 4$) than $u^2 + 7 v^2,$ although both are very well behaved. Both represent all odd primes $p \equiv 1,2,4 \pmod 7.$  Hmmm, probable correction for negative $k.$ It appears $x^2 + xy - 3 y^2$ and $u^2 - 13 v^2$ represent the same numbers. Good to know.
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