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.
On
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.
On
We can assume that $b\geq0$ and by your starting reasoning we ca get $b\in\{10,39,49\}$,
which gives all 12 solutions.
On
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.
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 ...