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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.

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    If you subtract $2bc$ from both sides, you get $(b-c)^2=2011-bc$. It's a beginning, you need more tricks to narrow down the possibilities, I think. Can $b$ and $c$ be negative as well as positive? – 2017-02-26
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    @HaraldHanche-Olsen: $(b+c)^2 - 2bc = b^2+c^2 \ne (b-c)^2$. – 2017-02-26
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    @HaraldHanche-Olsen If we replace $c\rightarrow -c$ then we can get $(b-c)^2=2011-bc$. And for 2nd one its not given in question. But when I checked this using WA they provide negative too, moreover they gave $12$ integral solutions. – 2017-02-26
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    @kennytm Oh, silly me. Meant to subtract $4bc$. – 2017-02-26
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    Check the pythagorean triples firmula. – 2017-02-26
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    Your work is correct. All you need to do is check integer values for $c$ in the interval $[-51,51]$, since the discriminant can be $\geq0$ and $c$ can be negative. The only integer values of $c$ that make $8044-3c^2$ an integer are $\pm10, \pm39, \pm49$. Plugging those values into the formula for $b$ gets the solution pairs. – 2017-02-26

4 Answers 4

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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 ...

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    beyond my level.... though, thanks for giving your time on this problem. – 2017-02-26
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    @mathlover Admittedly, yes. Incidentally, there is an even-more-beyond-level (for most) theory that would have told us right away "Oh, there is exactly one twelve-tuple of solutions ..:" from the fact that $2011$ is prime and $\equiv 1\pmod 3$. :) – 2017-02-26
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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.

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We can assume that $b\geq0$ and by your starting reasoning we ca get $b\in\{10,39,49\}$,

which gives all 12 solutions.

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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.