# Numbers integrally represented by a ternary cubic form

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 $f(a,b,c) \equiv 0 \pmod {q^3}.$ Checked correct for $q=3,5.$ Maybe I will do a few more.

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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Can $f$ represents large power with coprime $a,b,c$? In the bivariate case this is forbidden by an abc related conjecture. –  joro Apr 11 '13 at 13:02
@joro, I don't believe that will be a problem here, but I will do some checking. The main thing is that this behaves very much as the principal binary quadratic form of a discriminant, there is a rule for multiplication. –  Will Jagy Apr 11 '13 at 19:33
Sure, there are large powers with primitive representations. This follows from the multiplicativity that Will J. already noted. For instance, $f(3356898, 3732782, 5764967) = 7^{23}$ (I started from $f(2,1,0)=7$ and used $7^{23}=1^{10}7^{23}$ and $f(2,-1,0)=1$ to reduce the resulting $(a,b,c)=(472709258936428, 396738620092614, 257006830281609)$ ). –  Noam D. Elkies Apr 11 '13 at 20:32
@Noam, thanks. I've been checking, it appears the only primes to worry about are 2 and 11. I do think that the polynomial is not divisible by 8 unless all three variables are even ( a small finite check mod 8, not done yet). Perhaps something similar for 11, not sure yet. –  Will Jagy Apr 11 '13 at 21:12
$z^{k-1} x + y^k$ is surjective for all $k$ so large powers are represented primitively in arbitrary large degree with $z=1$. –  joro Apr 12 '13 at 8:16

Your conjectures are correct. So was the "someone else at MSRI [who] muttered something about norm forms" (mentioned in earlier edits of the question), except for the part about laughing at you.

As you in effect note, $f(a,b,c)$ is the norm $N_{K/{\bf Q}}(a+bx+cx^2)$, where $x$ is one of the roots of $x^3-x^2-x-1 = 0$ and $K$ is the cubic number field ${\bf Q}(x)$. This field has discriminant $-44$, and ${\bf Z}[x]$ is the full ring of integers $O_K$ (equivalently, the field discriminant of $K/{\bf Q}$ equals the polynomial discriminant of $x^3-x^2-x-1$; to check this in gp, compute

poldisc(x^3-x^2-x-1)
nfdisc(x^3-x^2-x-1)

and observe that both return $-44$). Now for (A), you already know that $x^3-x^2-x-1$ has at least one root modulo any prime $q$ unless $q$ is represented by the nonprincipal quadratic form $3u^2+2uv+4v^2$ of discriminant $-44$. (For other $q$: there's a triple root for $q=2$, a double and a simple root for $q=11$, three distinct roots for $q=u^2+11v^2$, and one simple root for odd $q$ not congruent to a square $\bmod 11$.) Equivalently, $K$ has an ideal of norm $q$ unless $q = 3u^2+2uv+4v^2$. But $O_K$ is a principal ideal domain, so once there's an ideal of norm $q$ then it has a generator $a+bx+cx^2 \in O_K$, and then $q=f(a,b,c)$ (or $q=f(-a,-b,-c)$ if we chose $a+bx+cx^2$ of norm $-q$). The discriminant of $K$ is small enough that one can check unique factorization by hand using the Minkowski bound; nowadays this exercise can also be done routinely on the computer, e.g. in gp

K = bnfinit(x^3-x^2-x-1); K.cyc

(This functionality happens to be one of the "Usage examples" in the current Wikipedia page on gp.)

[EDIT In fact this $K$ happens to be one of the handful of number fields whose Minkowski bound is so tight that nothing needs to be checked! The discriminant $\Delta_K = -44$ is small enough in absolute value that the bound $$\frac4\pi \frac{3!}{3^3} \left|\Delta_K\right|^{1/2} = 1.8768\ldots$$ is less than $2$, which means every ideal $I$ has a nonzero element of norm $\pm \left|I\right|$ and is thus automatically principal. TIDE]

(B) Translating the factorization of $x^3-x^2-x-1 \bmod q$ into the factorization of the ideal $(q)$ in $O_K$, we see that if $q = 3u^2+2uv+4v^2$ then $(q)$ remains prime in $O_K$, and thus that $q \mid N_{K/{\bf Q}}(a+bx+cx^2)$ iff $q \mid a+bx+cx^2$. For $q=2$ the ideal $(q)$ is the cube of $(1+x)$, so $8 \mid f(a,b,c)$ iff $a,b,c$ are all even. Any power of a prime $q$ other than those of the form $3u^2+2uv+4v^2$ can be represented primitively by $f$, even $q=11$ (for which $(q)$ factors as $(2+x)(3-2x)^2$). If we do not care about primitivity then we can also represent all powers of $2$, and all powers of $q^3$ for $q = 3u^2+2uv+4v^2$.

By multiplicativity this also proves the final conjecture: the nonzero $n \in {\bf Z}$ that are represented by $f$ are precisely those whose $q$-valuation is a multiple of $3$ for all primes $q = 3u^2+2uv+4v^2$.

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Thank you, Noam. –  Will Jagy Apr 11 '13 at 23:39
You're welcome, and thank you for accepting my answer. –  Noam D. Elkies Apr 12 '13 at 16:22
Oh, EDIT spelled backwards is TIDE thus indicating the end of an edited section... –  Will Jagy Apr 12 '13 at 18:16