I made this up years ago when Kap got interested in the "tribonacci" numbers. Later I asked someone else at MSRI about this, he muttered something about norm forms and laughed at me.
Given integers $a,b,c,$ and cubic form $$ f(a,b,c) = a^3 + b^3 + c^3 + a^2 b - a b^2 + 3 a^2 c - a c^2 + b^2 c - b c^2 - 4 a b c $$ $$ f(a,b,c) = \det \left( \begin{array}{ccc} a & b & c \\\ c & a + c & b + c \\\ b + c & b + 2 c & a + b + 2 c \end{array} \right) . $$ what primes $p$ can be integrally represented as $$ p = f(a,b,c)? $$
(A): I think it is all primes $(p| 11) = -1 ,$ and all $p = u^2 + 11 v^2$ in integers, but not any $q = 3 u^2 + 2 u v + 4 v^2.$ Note that, if $-p$ is represented, so is $p.$
(B): I also suspect that if prime $q = 3 u^2 + 2 u v + 4 v^2$ and $f(a,b,c) \equiv 0 \pmod q,$ then all three $a,b,c \equiv 0 \pmod q,$ and $f(a,b,c) \equiv 0 \pmod {q^3}.$
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 $$ X = \left( \begin{array}{ccc} 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ 1 & 1 & 1 \end{array} \right) $$ Then $X^3 = X^2 + X + I$ and $X^4 = 2 X^2 + 2 X + I.$
If all suspicions are correct, we can correctly describe all numbers integrally represented by this polynomial: positive or negative are unimportant, most prime factors are unimportant, all that matters is that every exponent of a prime factor $q = 3 u^2 + 2 u v + 4 v^2$ must be divisible by 3.
I should have done this last time: most of the class field part has already been done, by Hudson and Williams (1991), Theorem 1 and Table 1 on page 134. You get my version of the polynomial by negating their variable $x.$
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p a b c
2 0 1 1
7 0 -11 6
11 0 -3 2
13 0 -1 2
17 -1 0 2
19 1 2 4
29 0 -7 4
41 0 3 2
43 0 4 -1
47 0 5 -2
53 0 1 4
61 0 46 -25
73 2 -36 19
79 0 3 4
83 0 24 -13
101 -1 12 -6
103 0 15 -8
107 1 -9 5
109 1 2 6
127 1 -2 4
131 1 7 -3
139 1 -6 4
149 -1 4 2
151 0 -20 11
163 0 5 2
167 -1 1 5
173 0 6 -1
193 1 -52 28
197 0 9 -4
199 -1 5 1
211 0 -12 7
227 -2 0 5
233 0 -16 9
239 0 -6 5
241 0 -4 5
257 0 -1 6
263 2 4 9
269 -1 0 6
271 2 8 -3
277 1 -7 5
281 0 2 7
283 -1 2 6
293 -1 -8 6
307 2 -1 6
311 0 5 6
337 -2 5 2
347 1 7 5
349 0 19 -10
359 -1 9 -3
373 2 5 10
397 1 -1 7
401 0 -68 37
409 3 -77 41
419 0 -7 6
421 0 7 2
431 1 -14 8
439 0 8 -1
457 0 1 8
461 0 -2 7
479 1 -8 6
491 0 7 4
499 0 13 -6
503 -1 -36 20
523 0 9 -2
541 2 -12 7
547 1 -11 7
557 -1 25 -13
563 -2 -11 8
569 0 8 1
571 1 -3 7
587 0 -29 16
593 3 -25 13
599 -1 0 8
601 0 7 6
607 0 11 -4
613 0 4 9
617 2 -1 8
659 0 8 3
673 0 -6 7
677 0 -17 10
683 -1 4 8
701 2 13 -6
733 1 10 -2
739 -1 14 -6
743 -2 1 8
757 0 81 -44
761 -1 8 2
769 0 -25 14
773 -1 7 5
787 2 5 12
809 -1 -10 8
811 -4 0 7
821 -1 3 9
827 2 10 7
853 0 -11 8
857 -2 3 8
863 0 9 2
877 -2 -15 10
883 0 -14 9
887 2 -3 8
907 0 -5 8
911 0 8 7
919 0 -2 9
929 1 7 11
937 3 8 14
941 3 -1 9
953 -1 6 8
967 1 13 -5
991 1 -35 19
997 -3 7 3
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Primes represented by $x^2 + 11 y^2$ and then by $3 x^2 + 2 x y + 4 y^2,$ both up to $1000.$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
1 0 11
Discriminant -44
Modulus for arithmetic progressions?
11
Maximum number represented?
1000
p p mod 11
11 0
47 3
53 9
103 4
163 9
199 1
257 4
269 5
311 3
397 1
401 5
419 1
421 3
499 4
587 4
599 5
617 1
683 1
757 9
773 3
863 5
883 3
907 5
911 9
929 5
991 1
0 1 3 4 5 9
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
3 2 4
Discriminant -44
Modulus for arithmetic progressions?
11
Maximum number represented?
1000
p p mod 11
3 3
5 5
23 1
31 9
37 4
59 4
67 1
71 5
89 1
97 9
113 3
137 5
157 3
179 3
181 5
191 4
223 3
229 9
251 9
313 5
317 9
331 1
353 1
367 4
379 5
383 9
389 4
433 4
443 3
449 9
463 1
467 5
487 3
509 3
521 4
577 5
619 3
631 4
641 3
643 5
647 9
653 4
661 1
691 9
709 5
719 4
727 1
751 3
797 5
823 9
829 4
839 3
859 1
881 1
947 1
971 3
977 9
983 4
1 3 4 5 9
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
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joro asked about high powers being represented primitively. It turns out that the polynomial is not divisible by 8 unless $a,b,c$ are all even. This, despite the fact that 2 is represented. I believe this happens for all the (unrepresented) primes $q = 3 u^2 + 2 u v + 4 v^2$ as well, in the strongest manner: the polynomial is not divisible by $q$ itself unless $a,b,c$ are. I thought there might be trouble with the prime 11, but no. Anyway, here are some prime powers represented primitively, where $47 = 36 + 11$ and $53 = 9 + 44:$
7 0 1 2
49 1 -1 3
343 6 4 5
2401 -11 -3 9
16807 -11 30 -8
117649 -19 75 -29
823543 -2 -117 82
5764801 162 43 12
40353607 205 -64 186
11 -1 1 1
121 10 15 16
1331 -10 -2 7
14641 12 28 9
161051 1 25 59
1771561 53 -78 70
19487171 37 46 300
214358881 171 -210 460
13 1 3 3
169 10 17 18
2197 -4 3 10
28561 -15 -8 24
371293 8 71 34
4826809 -54 98 77
62748517 -257 125 167
47 1 3 5
2209 10 12 3
103823 108 181 202
4879681 104 32 153
229345007 -128 319 432
53 -1 1 3
2809 10 23 24
148877 100 163 170
7890481 100 18 187
418195493 342 -308 451
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