# Integers not represented by $2 x^2 + x y + 3 y^2 + z^3 - z$

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 $q = 2 u^2 + u v + 3 v^2.$

http://zakuski.math.utsa.edu/~jagy/inhom.html

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

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More succinctly, you're asking if it's true that a positive integer c is not of the form +-(2x^2+xy+3y^2+z^3-z) if (c,6)=1 and 27c^2-4 is 23 times a square? Is it even iff? What a great question! Looping with |x|,|y|,|z|<=100 I do indeed pick up every integer between -1000 and 1000 other than +-1 and +-599, and 1, 599 are precisely the two values of c less than 1000 and coprime to 6 satisfying the Pell equation. Of course I understand that you're asserting that you've computed much much further, I'm just (a) independently verifying a part of the calculations and (b) giving the problem a plug. – Kevin Buzzard Jan 21 '10 at 7:43
Much easier question: can someone show that N is of the form 2x^2+xy+3y^2+z^3-z iff -N is? Will: do you believe this? – Kevin Buzzard Jan 21 '10 at 7:43
Personally, I believe it is iff in your succinct version. But research about indefinite inhomogeneous polynomials has been rare. Usually there are positivity restrictions, see R.C.Vaughan book The Hardy-Littlewood Method.'' So I do not know of any theorem that says an indefinite homogeneous polynomial represents any number unless there is a good reason for failure'' but that is something I believe to be true. And I believe the +-N symmetry, which fails in the related problem for 3 x^2 + 2 xy + 4 y^2 + z^3 - z^2 - z because of the z^2 term. See website. or email me. Thanks for appreciation – Will Jagy Jan 21 '10 at 17:28
Will [or anyone else reading]: the reason this question got bumped was that I edited my answer. Initially I wrote "here's a strategy to prove iff and it should work" but I spent some time recently trying to finish the job, and failed. I edited my answer (which is now rather a mess :-/ ) to reflect this. The answer still answers the question you raise, but it does not do the other implication (it doesn't show that if C isn't a solution to the Pell then C is represented). See my edit after the statement of Conj 1 in my answer. – Kevin Buzzard Mar 29 '11 at 18:15

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.

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As for conjecture 2: the primes dividing D are the index divisors of the cubic polynomial defining the Hilbert class field. I know that common index divisors necessarily split; if this is true for all index divisors that are primes (except for finitely many divisors of the coefficients), Conj. 2 would follow since primes splitting in the Hilbert class field must be represented by the principal form. I guess there's some classical result on index divisors that takes care of Conj. 2. – Franz Lemmermeyer Jan 29 '10 at 14:28
Thank you, Kevin. – Will Jagy Jan 29 '10 at 18:21
@buzzard. I suggest that you also make a community wiki post with your mail to NMBRTHRY, and possible responses coming in there. – Anweshi Jan 29 '10 at 18:50
Thanks again, Kevin. – Will Jagy Feb 16 '10 at 19:09
Kevin, thanks for letting me know about the edit (I had one new comment when I logged in). And thanks for putting in some effort to finalize your presentation. As when you first suggested the idea, I remain fascinated by the possibility of proving that all other numbers are integrally represented by the polynomial. Do you think you can do the difficult direction for the (much) easier polynomial in my Problem in the December 2010 M.A.A. Monthly? Of course I did not ask the difficult direction as part of the Problem, but have been hoping people might include that in their submitted solutions. – Will Jagy Mar 29 '11 at 20:59