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Fact. Let theta be a zero of X^3-X+1, let eta in Z[theta] be a zero of X^3-X+C with C in Z odd, and let p be a prime number that is inert in Z[theta]. Then p does not divide index(Z[theta]:Z[eta]).

Proof. By hypothesis, Z[theta]/pZ[theta] is a field of size p^3. Let e be the image of eta in that field. Since X^3-X+C is irreducible in Z[X] (even mod 2), it is the characteristic polynomial of eta over Z. Hence its reduction mod p is the characteristic polynomial of e over Z/pZ. If now e is in Z/pZ, then that characteristic polynomial also equals (X-e)^3, so that in Z/pZ we have 3e = 0 and 3e^2 = -1, a contradiction. Hence e is not in Z/pZ, so (Z/pZ)[e] = Z[theta]/pZ[theta], which is the same as saying Z[theta] = Z[eta] + pZ[theta]. Then p acts surjectively on the finite abelian group Z[theta]/Z[eta], so the order of that group is not divisible by p. End of proof.

(EDIT: dollar signs added - GM)

Fact. Let $\theta$ be a zero of $X^3-X+1$, let $\eta$ in ${\bf Z}[\theta]$ be a zero of $X^3-X+C$ with $C$ in $\bf Z$ odd, and let $p$ be a prime number that is inert in ${\bf Z}[\theta]$. Then $p$ does not divide index$({\bf Z}[\theta]:{\bf Z}[\eta])$.

Proof. By hypothesis, ${\bf Z}[\theta]/p{\bf Z}[\theta]$ is a field of size $p^3$. Let $e$ be the image of $\eta$ in that field. Since $X^3-X+C$ is irreducible in ${\bf Z}[X]$ (even mod 2), it is the characteristic polynomial of $\eta$ over $\bf Z$. Hence its reduction mod $p$ is the characteristic polynomial of $e$ over ${\bf Z}/p{\bf Z}$. If now $e$ is in ${\bf Z}/p{\bf Z}$, then that characteristic polynomial also equals $(X-e)^3$, so that in ${\bf Z}/p{\bf Z}$ we have $3e = 0$ and $3e^2 = -1$, a contradiction. Hence $e$ is not in ${\bf Z}/p{\bf Z}$, so $({\bf Z}/p{\bf Z})[e] = {\bf Z}[\theta]/p{\bf Z}[\theta]$, which is the same as saying ${\bf Z}[\theta] = {\bf Z}[\eta] + p{\bf Z}[\theta]$. Then $p$ acts surjectively on the finite abelian group ${\bf Z}[\theta]/{\bf Z}[\eta]$, so the order of that group is not divisible by $p$. End of proof.

So I am claiming Conj 2 implies Conj 1. I don't know how to prove Conj 2 but it looks very accessible. Note that the Pell equation is related to units in $\mathbf{Q}(\sqrt{69})$ and the $2x^2+xy+3y^2$ is related to factorization in $\mathbf{Q}(\sqrt{-23})$. I've seen other results relating the arithmetic of $\mathbf{Q}(\sqrt{D})$ and $\mathbf{Q}(\sqrt{-3D})$.

Ok, so assuming Conjecture 2, let me sketch a proof of Conjecture 1.

Once this is done properly we have a solution to $x^2+xy+6y^2=z^3-z+C$, so we have written $z^3-z+C$ as the norm of a principal ideal in the integers of $\mathbf{Q}(\sqrt{-23})$. What we need to do now is to write it as the norm of a non-principal ideal, and of course we'll be able to do this if we can find some prime $p$ dividing $z^3-z+C$ which splits in $\mathbf{Q}(\sqrt{-23})$ into two non-principal primes, because then we replace one of the prime divisors above $p$ in our ideal by the other one. What we need then is to show that if the discriminant of $z^3-z+C$ is not $-23$ times a square, then there is some prime $p$ of the form $2x^2+xy+3y^2$ dividing some number of the form $z^3-z+C$ which is the norm of a principal ideal. This should follow from the Cebotarev density theorem, because Mordell's methods construct a huge number of solutions to $x^2+xy+6y^2=z^3-z+C$ which are "only constrained modulo 23", and so one should presumably be able to find a prime which splits in $\mathbf{Q}(\sqrt{-23})$, splits completely in the splitting field of $z^3-z+C$ and doesn't split completely in the splitting field of $z^3-z+1$. I have run out of energy to deal with this point however, so again there is a hole here. This issue seems analytic to me, and I am not much of an analytic guy.

[EDIT/clarification: Jagy only asks one direction of the iff in his question, and this answer below gives a complete answer to the question Jagy asks. I came back to this question recently though [I am writing this para a year after I wrote the original answer] and tried to fill in the details of the argument in the other direction (proving that if C was not an odd integer solution to $27C^2-4=23D^2$ then $C$ was represented by the form) and I failed. So the "hole" I flag in the answer below still really is a hole, and this post still remains an answer to Jagy's question, but not a complete proof of Conjecture 1, which should still be regarded as open.]

So I am claiming Conj 2 implies the "only if" version of Conj 1. I don't know how to prove Conj 2 but it looks very accessible [edit: I do now; see below]. Note that the Pell equation is related to units in $\mathbf{Q}(\sqrt{69})$ and the $2x^2+xy+3y^2$ is related to factorization in $\mathbf{Q}(\sqrt{-23})$. I've seen other results relating the arithmetic of $\mathbf{Q}(\sqrt{D})$ and $\mathbf{Q}(\sqrt{-3D})$.

Ok, so assuming Conjecture 2, let me sketch a proof of the "only if" part of Conjecture 1.

Once this is done properly we have a solution to $x^2+xy+6y^2=z^3-z+C$, so we have written $z^3-z+C$ as the norm of a principal ideal in the integers of $\mathbf{Q}(\sqrt{-23})$. What we need to do now is to write it as the norm of a non-principal ideal, and of course we'll be able to do this if we can find some prime $p$ dividing $z^3-z+C$ which splits in $\mathbf{Q}(\sqrt{-23})$ into two non-principal primes, because then we replace one of the prime divisors above $p$ in our ideal by the other one. What we need then is to show that if the discriminant of $z^3-z+C$ is not $-23$ times a square, then there is some prime $p$ of the form $2x^2+xy+3y^2$ dividing some number of the form $z^3-z+C$ which is the norm of a principal ideal. This should follow from the Cebotarev density theorem, because Mordell's methods construct a huge number of solutions to $x^2+xy+6y^2=z^3-z+C$ which are "only constrained modulo 23", and so one should presumably be able to find a prime which splits in $\mathbf{Q}(\sqrt{-23})$, splits completely in the splitting field of $z^3-z+C$ and doesn't split completely in the splitting field of $z^3-z+1$. I have run out of energy to deal with this point however, so again there is a hole here. This issue seems analytic to me, and I am not much of an analytic guy. [edit: I came back to this question a year later and couldn't do it, so this should not be regarded as a proof of the "if" part of Conj 1]

EDIT: OK so here, verbatim, is an email from Lenstra in which he establishes Conjecture 2. Note: when he says "2)" he means the statement that if C is an odd solution to the Pell and if theta is a root of $X^3-X+1=0$ then $\mathbf{Z}[\theta]$ contains a root of $X^3-X+C$ (I sketched a proof of this above).

Let eta in Z[theta] be a zero of X^3-X+C. Then |D| = index(Z[theta]:Z[eta]). Now let p be a prime number of the form 2x^2+xy+3y^2. To be proved: p does not divide D. By class field theory over Q(sqrt(-23)), the prime p is inert in Z[theta], so Z[theta]/pZ[theta] is a field of size ppp. Look at the image e (say) of eta in that field. It is nonzero, since from eta's equation it is clear that eta/p is not an algebraic integer. First case: e is in the prime field. Then since the trace of eta is 0, the trace 3e of e is zero but e isn't, so p = 3. And 3 visibly doesn't divide D by 27CC-4=23DD. (Actually, if we believe 2), then this first case cannot occur, since e in prime field implies Z[eta] subset Z + pZ[theta] hence 3|index = |D| after all.) Second case: e is not in the prime field. Then e generates the entire field, so Z[theta] = Z[eta] + pZ[theta], so p acts surjectively on the finite abelian group Z[theta]/Z[eta], so its order |D| is not divisible by p. End of proof.

EDIT: OK so here, verbatim, is an email from Lenstra in which he establishes Conjecture 2.

Fact. Let theta be a zero of X^3-X+1, let eta in Z[theta] be a zero of X^3-X+C with C in Z odd, and let p be a prime number that is inert in Z[theta]. Then p does not divide index(Z[theta]:Z[eta]).

Proof. By hypothesis, Z[theta]/pZ[theta] is a field of size p^3. Let e be the image of eta in that field. Since X^3-X+C is irreducible in Z[X] (even mod 2), it is the characteristic polynomial of eta over Z. Hence its reduction mod p is the characteristic polynomial of e over Z/pZ. If now e is in Z/pZ, then that characteristic polynomial also equals (X-e)^3, so that in Z/pZ we have 3e = 0 and 3e^2 = -1, a contradiction. Hence e is not in Z/pZ, so (Z/pZ)[e] = Z[theta]/pZ[theta], which is the same as saying Z[theta] = Z[eta] + pZ[theta]. Then p acts surjectively on the finite abelian group Z[theta]/Z[eta], so the order of that group is not divisible by p. End of proof.