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Expired by this question Show determinant of matrix is non-zero I am moved to ask:

Given integers $a,b,c,$ and cubic form $$ f(a,b,c) = a^3+2 b^3-6 a b c+4 c^3 = \left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right|, $$ what primes $p$ can be integrally represented as $$ p = f(a,b,c)? $$

I think it is $3,$ all primes $p \equiv 2 \pmod 3,$ and all $p = u^2 + 27 v^2$ in integers, but not any $q = 4 u^2 + 2 u v + 7 v^2.$ I checked for $p < 10000.$

Note that, if $-p$ is represented, so is $p.$

Although it does not finish things, note that if $f$ integrally represents both $m,n$ then it represents $mn.$ That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where $$ X = \begin{bmatrix} 0 & 0 & 2\\1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix}. $$ Then $X^3 = 2 I$ and $X^4 = 2 X.$

I once asked a guy at MSRI about pretty much the same problem, only instead of the important polynomial being $\lambda^3 - 2$ it was $\lambda^3 - \lambda^2 - \lambda - 1.$ The phrase norm forms came up, and he laughed at me.

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   p           a           b           c
   2           0           1           0
   3          -1           0           1
   5           1           0           1
  11          -1           1           1
  17          -1          -1           2
  23           1           1           2
  29          -3          10          -6
  31          -1          26         -20
  41           1          -2           2
  43           1          -1           2
  47          -1           4          -2
  53           1          -4           3
  59           1           3          -2
  71          -1           2           2
  83           3           1           3
  89           1           2           3
 101           3          -7           4
 107          -1           0           3
 109           1         -12           9
 113           1           4           2
 127          -1          16         -12
 131           3           3          -1
 137          -3           1           3
 149           1           4          -1
 157          -1           5          -2
 167          -3           3           2
 173          -3           7          -3
 179           1         -31          24
 191          -1          -2           4
 197           5           2          -1
 223           1           5           2
 227           3          -2           3
 229          -1          -1           4
 233           1           5          -3
 239           1           3           4
 251          -1          -4           5
 257           1           0           4
 263           3           4          -3
 269          -1           9          -6
 277           1           5          -1
 281          -1           1           4
 283          -1           8          -5
 293           1          -9           7
 307          -1           4           3
 311           3           3           5
 317          -3           5           1
 347           3         -12           8
 353           3          -1           4
 359          -5          23         -15
 383          -5          28         -19
 389          -3           2           4
 397           1           7          -5
 401           1          -5           5
 419           3           6           5
 431           1          -7           6
 433          -1          -5           6
 439           3           1           5
 443           3          -4           4
 449           1           8          -6
 457           1           2           5
 461           5           4          -2
 467          -1          -1           5
 479          -1           4           4
 491           3          18         -16
 499          -1           0           5
 503           5           3           6
 509           1           4           5
 521           5           5          -1
 557          -1          89         -70
 563           3           6          -1
 569          -1           7          -2
 587           3           4           6
 593          -7           2           5
 599           1           7          -4
 601           1         -22          17
 617          -5         -59          50
 641           3          23         -20
 643          -1           3           5
 647          -1          14         -10
 653           1          16         -13
 659           3         -10           7
 677          -1         -11          10
 683          -5           5           3
 691           3          -2           5
 701          -1          -3           6
 719           5           5          -4
 727           3          -5           5
 733           3           9          -8
 739           1           7          -2
 743           3          -3           5
 761           1         -14          11
 773           5          -1           5
 797          -3           3           5
 809           1           2           6
 811           1           3           6
 821           3           7           6
 827          -1          11          -7
 839          -3           5           4
 857          -9           5           4
 863          -1           0           6
 881           1          -4           6
 887           7           3           7
 911          -1          -5           7
 919          -1           7           3
 929           9           2          -2
 941           9           3          -1
 947          -3           1           6
 953          -7          26         -16
 971          -1           8          -1
 977           1           7           5
 983           3           7          -3
 997           3         -11           8

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

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  • 1
    $\begingroup$ Expired or inspired? :) $\endgroup$ – Amzoti Mar 14 '13 at 2:23
  • $\begingroup$ @Amzoti, I think expired gives the more accurate feel. $\endgroup$ – Will Jagy Mar 14 '13 at 2:23
  • $\begingroup$ I can't tell you how many times I've been there! :) + 1 for a very interesting question! $\endgroup$ – Amzoti Mar 14 '13 at 2:24
  • $\begingroup$ Is there a path to go from $p|f(a,b,c)$ for some $a,b,c$ to $p = f(a,b,c)$ for some $a,b,c$? $\endgroup$ – user27126 Mar 14 '13 at 4:50
  • $\begingroup$ @Sanchez, always possible. The situations I know where that argument works are quadratic forms rather than cubic. $\endgroup$ – Will Jagy Mar 14 '13 at 5:02
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The discriminant of $\Bbb Q \subset \Bbb Q(\sqrt[3]2)$ is $-108$, and the Minkowski bound for this extension is $\frac {3!}{3^3} \frac 4 \pi \sqrt {108} \approx 2.940$. So to prove that this number field has class number $1$ we only need to find a way to represent $2$, and $2$ is indeed represented by $(0,1,0)$. Thus $p$ is represented by this norm form if and only if the ideal $(p)$ has an ideal factor of norm $p$, which happens if and only if $2$ is a cube modulo $p$.

If $p \equiv 2 \pmod 3$ then any nonzero element of $\Bbb F_p$ has one cube root in $\Bbb F_p$ and two cube roots in $\Bbb F_p^2$, so $2$ is a cube modulo $p$.

If $p \equiv 1 \pmod 3$ then $(p)$ splits in $\Bbb Q(\zeta_3)$, and $2$ is a cube if and only it further splits in $\Bbb Q(\zeta_3,\sqrt[3]2)$. Since $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$ is an abelian extension, it is a ray class field for some modulus $\mathfrak m$ of $\Bbb Q(\zeta_3)$.

Working modulo $6$, we have $(a+b\zeta_3)^3 = (a^3+b^3) - 3ab^2+3ab(a-b)\zeta_3 = a^3 + b^3-3ab^2 \in \Bbb Z/6 \Bbb Z$, and thus for any $a,b,c \in \Bbb Z[\zeta_3]$, $a^3+2b^3+4c^3-6abc = a^3+2b^3+4c^3 \in \Bbb Z/6\Bbb Z$. So, norms that are coprime to $6$ are units ($\pm 1$) modulo $6$. So $\Bbb Q(\zeta_3,\sqrt[3]2)$ is an extension of the ray class field of modulus $(6)$ for $\Bbb Q(\zeta_3)$.

On the other hand, $G = (\Bbb Z[\zeta_3]/(6))^*/\langle \overline{\zeta_6} \rangle$ is isomorphic to $\Bbb Z/3 \Bbb Z$, which is the Galois group of the extension $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$, so $\mathfrak m = (6)$, and $2$ is a cube modulo $p$ if and onlt if $p \equiv 2 \mod 3$ or $p = a^2-ab+b^2$ where $a+\zeta_3 b$ is congruent modulo $6$ to one of $\{1,1+\zeta_3,\zeta_3,-1,-1- \zeta_3,- \zeta_3\}$.

Each element of $G$ (modulo complex conjugation) corresponds to a class of primitive binary quadratic forms of discriminant $-108$, or a corresponding lattice class (modulo multiplication by a unit and complex conjugation) whose endomorphism ring is $\Bbb Z[3\sqrt{-3}]$:

$\Lambda = \langle 1, 3\sqrt{-3} \rangle$ is a lattice corresponding to the neutral element of $G$ : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the neutral class.
while $\Lambda = \langle 2, \frac {1+3\sqrt{-3}}2 \rangle$ corresponds to the other two classes : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the other two classes.

So if $p \equiv 1 \pmod 3$, then $p$ is represented either as $a^2 + 27b^2$ (when $2$ is a cube) or as $4u^2 \pm 2uv + 7v^2$ (when $2$ is not a cube), and never both at the same time.

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  • $\begingroup$ Minor comment: the criterion in the first paragraph, that $p$ is a norm iff it is not prime in the cubic field, requires noting that $-1$ is itself a norm so that there is no need to distinguish between $-p$ or $p$ being a norm; they both are or both aren't. If $-1$ is not a norm in a quadratic field, for instance, then $-p$ may be a norm of some algebraic integer while $p$ is not. $\endgroup$ – KCd Mar 20 '13 at 12:44
  • $\begingroup$ @KCd and mercio, Thank you. I just posted the original question as math.stackexchange.com/questions/336191/… but his time I put the source (Hudson and Williams) I should have posted with this question. The area where I would enjoy additional tutoring is "Thus $p$ is represented by this norm form if and only if the ideal $(p)$ is not prime in this extension" $\endgroup$ – Will Jagy Mar 20 '13 at 19:51
  • $\begingroup$ @WillJagy : it goes along these lines : there is an element of norm $p$, iff. there is a principal ideal of norm $p$, iff. (class number $1$) there is an ideal of norm $p$ (which necessarily contains the ideal $(p)$ of norm $p^3$), iff. there is a factorisation $(p) = (\mathfrak P_1)(\mathfrak P_2)$ with $N(\mathfrak P_1) = p$, iff. $X^3-2$ has a linear factor modulo $p$, iff. $2$ is a cube modulo $p$ $\endgroup$ – mercio Mar 20 '13 at 20:23
  • $\begingroup$ @mercio, thank you. I simply never studied algebraic number theory, but in the natural course of things made up some problems where that is the method of choice, some still open. $\endgroup$ – Will Jagy Mar 20 '13 at 20:46

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