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For a prime number $p \neq 11$, do we have $p = x^2 + xy + 3y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1, 3, 4, 5, 9$ mod $11$?

An example where this is true:$$5 = 1^2 + 1 \times 1 + 3 \times 1^2.$$I know that $\mathbb{Z}[(1 + \sqrt{-11})/2]$ is a PID.

user26857
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    Have you tried any of the techniques from your other very similar question? In particular, my answer to that question works in exactly the same way, replacing $X^2+3$ with $X^2+11$. – Mathmo123 Apr 01 '16 at 20:13
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    Moreover, I think it's quite rude that you asked a question, received 8 answers, didn't comment on/accept any of them, and didn't at least attempt to use them to answer this question, when the techniques from any of them would've worked here too. – Mathmo123 Apr 01 '16 at 20:25

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If $(p| 11) = 1,$ then the Legendre symbol $(-11|p) = 1.$ We can solve $\beta^2 \equiv -11 \pmod p.$ By choosing either $\beta$ or $p - \beta,$ because this time we need $\beta$ odd, we can solve $\beta^2 \equiv -11 \pmod {4p}.$ That is, $\beta^2 = -11 + 4pt$ for some integer $t.$

So far, we have the binary quadratic form $\langle p, \beta, t \rangle,$ or $$ f(x,y) = px^2 + \beta xy + t y^2, $$ of discriminant $-11.$

We apply Gauss reduction to get a reduced form; inequalies show that the only reduced form of discriminant $-11$ is $\langle 1, 1, 3 \rangle.$ The 2 by 2 integer matrix $P$ of determinant $1$ that took us from $\langle p, \beta, t \rangle$ to $\langle 1, 1, 3 \rangle$ has an inverse of integers, $P^{-1}.$ The left hand column of $P^{-1}$ shows how to represent $p$ as $x^2 + xy + 3y^2.$

Let's see; we find $P$ one step at a time. In the end, though, we have $$ P^T G P = H, $$ where $$ G = \left( \begin{array}{rr} p & \frac{\beta}{2} \\ \frac{\beta}{2} & t \end{array} \right) $$ and $$ H = \left( \begin{array}{rr} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{array} \right) $$ If we name $Q = P^{-1},$ we have $$ Q^T H Q = P. $$

Will Jagy
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  • Given that you have more or less been able to copy your answer to the OP's other question, I think this question counts as a duplicate. – Mathmo123 Apr 01 '16 at 20:18
  • @Mathmo123, yes. Not entirely sure why I bothered, unless somebody actually reads this one because I posted early. This method does use the least, um, technology, if one accepts quadratic reciprocity as given. It also gives a construction. – Will Jagy Apr 01 '16 at 20:25
  • @Mathmo123 I see, this is your dissertation area. If you want something really difficult, prove both directions in an if and only if for What integers $n$ are expressible as $4x^2 + 2 xy + 7 y^2 - z^3,$ with $x,y,z$ integers with no sign restrictions. Similar question can be constructed from any discriminant of positive binary quadratic forms of class number three. In comparison, $x^2 + 27 y^2 - z^3$ is universal. – Will Jagy Apr 01 '16 at 20:31
  • @WillJagy thanks. I'll have a think. The $z^3$ term is throwing me off balance a bit. – Mathmo123 Apr 01 '16 at 20:46
  • @Mathmo123, finding the pattern of numbers not represented is easy on computer, and proving they are truly not represented is not so bad. The hard part is showing everything else really is represented. I did put this as a question here, http://math.stackexchange.com/questions/255834/what-numbers-are-integrally-represented-by-4-x2-2-x-y-7-y2-z3 – Will Jagy Apr 01 '16 at 20:54
  • @Mathmo123 evidently I asked that one on MO first http://mathoverflow.net/questions/104889/what-numbers-are-integrally-represented-by-4-x2-2-x-y-7-y2-z3 – Will Jagy Apr 01 '16 at 21:00