EDIT, 9 March 2014: when I asked this in 2010, I did not have the courage of my convictions, and so did not ask for an if and only if proof, as Kevin Buzzard quite properly pointed out. Such problems are now somewhat known open problems, as I told them to some experts in 2011. Probably the easiest of the bunch: It is easy to describe a set of integers that are not represented by $4 x^2 + 2 x y + 7 y^2 - z^3;$ I even sent that in as a Monthly problem (December 2010, problem 11539), only one solver, Robin Chapman(December 2012). **Open problem**: can we prove that the polynomial integrally represents every other number? There is a similar open problem for each discriminant of positive binary quadratic forms with class number three, including the other direction for Kevin Buzzard's answer below.


ORIGINAL: The following problem is my variant of something Irving Kaplansky noticed when we worked together. I do not think it is by nature a difficult problem, it is simply too hard for me to finish.


Suppose we have an integer $C > 0$ that is not divisible by 2 or 3, while there is another integer $F > 0$ such that 
$ 27 C^2 - 23 F^2 = 4.$


For any integers $x,y,z,$ 
is it true that $ 2 x^2 + x y + 3 y^2 + z^3 - z \neq C $ and
 $ 2 x^2 + x y + 3 y^2 + z^3 - z \neq -C ,$ or together
 $ 2 x^2 + x y + 3 y^2 + z^3 - z \neq \pm C $ ?


I have proved it for the four smallest values of $C,$ those being 
 1, 599, 14951, 9314449. The case $C=1$ comes directly from the 
Hudson and Williams paper below. Note that the even values of $C$ fail miserably, they seem to all be values of $ z^3 - z.$ The polynomial $ g(x,y,z) = 2 x^2 + x y + 3 y^2 + z^3 - z $
represents every other number $n$ with 
$ -10,000,000 \leq n \leq 10,000,000,$ according to my computer.


The Spearman and Williams article (see below) is explicit class field theory, not a topic I know.
I should point out that, in retrospect, what I proved for the four smallest (odd) $C$ seems to amount to the statement that $z^3 - z + C$ is irreducible $\pmod q$ for any prime $ q = 2 u^2 + u v + 3 v^2.$


(27 January 2010) I finally got smart and decided to do a part check of the "irreducible" version. For the next three values of $C,$ those being 232488049, 144839681351, 3615189146999, I factored $z^3 - z + C \pmod q$ and found it to be irreducible for primes $q < 1000$ and $ q = 2 u^2 + u v + 3 v^2.$


The two main references are:


Blair K. Spearman and Kenneth S. Williams,
 "The Cubic Conguence $x^3 + {A} x^2 + {B} x + {C} \equiv 0 \bmod p $
 and Binary Quadratic Forms",
 Journal of the London Mathematical Society,
 volume 46,
 1992,
 pages 397-410


Richard H. Hudson and Kenneth S. Williams",
 "Representation of primes by the principal form of
 discriminant $-{D}$ when the classnumber $h(-{D})$ is $3$",
 Acta Arithmetica,
 volume 57,
 1991,
 pages 131-153.


One needs this Lemma: if an integer $n$ has an integer representation as 
$ n = 2 x^2 + x y + 3 y^2,$ then $n$ is divisible by some prime 
$ q = 2 u^2 + u v + 3 v^2.$


Everything I know about this problem is in pdf's at (Feb. 2018):


<http://zakuski.utsa.edu/~jagy/inhom.cgi>


including a proof of the preceding Lemma in jagy\_division.pdf 
and the proof for the four smallest $C$ in jagy\_conjecture\_23.pdf
and a list of intimately related problems in jagy\_list.pdf
.


I welcome individual responses to this along with posted answers or comments. One of my email addresses can be found using the search feature at 


<http://www.ams.org/cml>



Integers not represented by $ 2 x^2 + x y + 3 y^2 + z^3 - z $
EDIT: Hendrik Lenstra emailed me a proof of Conjecture 2. I'll append it below. So Jagy's question is now solved.




---


OK so I think that Jagy wants to make the following conjecture:


CONJECTURE 1: an integer $C$ is not representable by the form F(x,y,z)=2x^2+xy+3y^2+z^3-z if, and only if, $C$ is odd and $27C^2-4=23D^2$ with $D$ an integer.


[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.]


I have a proof strategy for this. I am too lazy to fill in some of the details though, so maybe a bit of it doesn't work, but it should be OK. However, I am also reliant on a much easier-looking conjecture (which I've tested numerically so should be fine, but I can't see why it's true):


CONJECTURE 2: if $C$ is odd and $27C^2-4=23D^2$, then there's no prime p
dividing D of the form $2x^2+xy+3y^2$.


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.


The Pell equation is intimately related to the recurrence relation


$$t\_{n+2}=25t\_{n+1}-t\_n$$


with various initial conditions. For example the positive $C$s which
are solutions to $27C^2-4=23D^2$ are all generated by this recurrence
starting at $C\_1=C\_2=1$, and the $D$s are all generated by the same
recurrence with $D\_1=-1$ and $D\_2=1$. Note that $C\_n$ is even iff $n$
is a multiple of 3, and (by solving the recurrence explicitly) one
checks easily that $C\_{3n}=(3C\_{n+1})^3-(3C\_{n+1})$, so we've represented
the even solutions to the Pell equation as values of $F$ (with $x=y=0$).


Let's then consider the odd solutions to the Pell equation. Say $C$
is one of these. We want to prove that there is no solution in
integers $x,y,z$ to 


$$2x^2+xy+3y^2=z^3-z+C.$$


Let's do it by contradiction. Consider the polynomial $Z^3-Z+C$. First
I claim it's irreducible. This is because it is monic, of degree 3,
and has no integer root, because $C$ is odd. Next I claim that
the splitting field contains $\mathbf{Q}(\sqrt{-23})$. This is
because of our Pell assumption and the fact that the discriminant
of $Z^3-Z+C$ is $4-27C^2$. Next I claim that the splitting
field of $Z^3-Z+C$ is in fact the Hilbert class field of
$\mathbf{Q}(\sqrt{-23})$. I only know an ugly way of seeing this:
if $\theta$ is a root of $Z^3-Z+1=0$ then I know recurrence relations
$e\_n$, $f\_n$ and $g\_n$ (all defined using the relation above but with
different initial conditions) with $e\_n\theta^2+f\_n\theta+g\_n$ a root of
$Z^3-Z+C\_{3n+1}$, and other relations giving roots of $Z^3-Z+C\_{3n+2}$
and $Z^3-Z-C\_{3n+1}$ and $Z^3-Z-C\_{3n+2}$. Most unenlightening but it
does the job because it embeds $\mathbf{Q}(\theta)$ into the splitting
field, and the Galois closure of $\mathbf{Q}(\theta)$ is the Hilbert
class field of $\mathbf{Q}(\sqrt{-23})$. 


Right, now for the contradiction, assuming Conjecture 2. Let's assume
that $C$ is a solution to the Pell, and $z^3-z+C$ can be written $2x^2+xy+3y^2$.
Now $C$ is odd so $z^3-z+C$ isn't zero, and hence it's positive,
so it's the norm of a non-principal ideal~$I$ in the integers $R$ of
$\mathbf{Q}(\sqrt{-23})$. This ideal $I$ is a product of prime ideals,
and $I$ isn't principal, so one of the prime ideals had better also not
be principal. Say this prime ideal has norm $p$. We conclude that $p$
divides $z^3-z+C$ and $p$ is of the form $2x^2+xy+3y^2$. Note in
particular that this implies $p\not=23$. Also $p\not=3$, because $C$
is odd and (because of general Pell stuff) hence prime to 3.


CASE 1: $p$ is coprime to $D^2$ (with $27C^2-4=23D^2$). In this
case the polynomial $Z^3-Z+C$ has non-zero discriminant mod $p$
(because $p\not=23$) and furthermore has a root $Z=z$ mod $p$.
Hence mod $p$ the polynomial either splits as the product of a linear
and a quadratic, or the product of three linears. This tells us
something about the factorization of $p$ in the splitting field
of $Z^3-Z+C$: either $p$ remains inert in $\mathbf{Q}(\sqrt{-23})$,
or it splits into 6 primes in the splitting field and hence splits
into two principal primes in $\mathbf{Q}(\sqrt{-23})$ (because the
principal primes are the ones that split completely in the Hilbert
class field). In either case $p$ can't be of the form $2x^2+xy+3y^2$,
so this case is done.


CASE 2: This is simply Conjecture 2.


In both cases we have our contradiction, and
so we have proved, so far, assuming Conjecture 2, that a solution $C$ to $27C^2-4=23D^2$
is representable as $2x^2+xy+3y^2+z^3-z$ iff it's even. 


Note that Conjecture 2 can be verified by computer for explicit values
of $C$, giving unconditional results---for example I checked in just
a few seconds that any odd $C$ with $|C|<10^{72}$ and satisfying the
Pell equation was not representable by the form, and that result
does not rely on anything. At least that's something concrete for Jagy.




---


OK so what about the other way: say $27C^2-4$ is not 23 times a square.
How to go about representing $C$ by our form? Well, here I am going to
be much vaguer because there are issues I am simply too tired to deal
with (and note that this is not the question that Jagy asked anyway).
Here's the idea. Look at the proof of Theorem 2 in Jagy's pdf Mordell.pdf.
Here Mordell gives a general algorithm to represent certain integers
by (quadratic in two variables) + (cubic in one variable). If you
apply it not to the form we're interested in, but to the following
equation:


$$x^2+xy+6y^2=z^3-z+C$$


then, I didn't check all the details, but I convinced myself that they
could easily be checked if I had another hour or two, but I think that
the techniques show that whatever the value of $C$ is, this equation
has a solution. The idea is to fix $C$, let $\theta$ be a root of
the cubic on the right (which we can assume is irreducible, as if it
were reducible then we get a solution with $x=y=0$), to rewrite the right
hand side as $N\_{F/\mathbf{Q}}(z-\theta)$, with $F=\mathbf{Q}(\theta)$
and now to try and write $z-\theta$ as $G^2+GH+2H^2$ with
$G,H\in\mathbf{Z}[\theta]$. Mordell does this explicitly (in a slightly
different case) in the pdf. The arguments come out the same though,
and we end up having to check that a certain cubic in four variables
has a solution modulo~23 with a certain property. I'll skip the painful
details. The cubic depends on $C$ mod 23, and so a computer calculation
can deal with all 23 cases.


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. 


(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.


