By computer calculations, I found the following conjecture that the quadratic form $4x^2 + 2xy + 3y^2 + 4w^2 + 2wz + 3z^2$ represents all primes except for the two primes 2 and 11. Is it possible to prove the conjecture? Or, are there results to attack the conjecture?

1$\begingroup$ what is the smallest prime which you have not been able to represent in this form? $\endgroup$– Yemon ChoiFeb 19, 2012 at 8:02

2$\begingroup$ @Yemon, it is almost certain that the form represents all primes except $2,11$ and all squarefree numbers except $1,2,11,22.$ $\endgroup$– Will JagyFeb 19, 2012 at 9:14

$\begingroup$ edited only to put the formula for the quadratic form in TeX. $\endgroup$– Noam D. ElkiesFeb 20, 2012 at 21:10
3 Answers
I don't know how much to say. Your form is in the same genus as $w^2 + x^2 + 11 y^2 + 11 z^2.$ It is isotropic in the 2adic numbers but anisotropic in the 11adic numbers. It is extremely likely that your form represents all positive integers except $1,11,121, 1331, \ldots$ and $2,22,242, 2662, \ldots,$ that is $11^k, 2 \cdot 11^k.$ Anisotropic means that your form integrally represents a number $n$ if and only if it represents $121 n.$ So, in opposition to any positive binary form, your quaternary form represents a set of positive density in the integers, indeed probably of full density.
I do not see a regular ternary form represented by yours. This rules out one method for proving the assertion I make above. However, that may not be the end of the story. In any case, it typically takes me a few days to finish one of these arguments.
If there is no clean proof, there is another direction. Manjul Bhargava and Jonathan Hanke, in something called the 290 Theorem, use analytic methods to show that certain forms in at least four variables represent all numbers above some large bound. What is unusual is that the bounds here are effective, which is not possible in three variables. For this reason, the new preprint by Jeremy Rouse does not finalize his project, although he does show that the natural conjecture follows from an extended Riemann Hypothesis.
Note that, for our amusement, $w^2 + x^2 + 11 y^2 + 11 z^2$ seems to represent all positive integers except $11^k \cdot \{3,6,7,14 \}$
Oh, almost forgot. Both forms represent a subset of the numbers represented by $$ x^2 + x y + 3 y^2 + w^2 + w z + 3 z^2, $$ which actually represents all positive integers. Your form is the one just above with my $x,w$ even, while $(1,1,11,11)$ is the form just above with $y,z$ even, but then "reduced."
Enough for now. I will post what I am able to prove later.
EDIT, Sunday, 19 February 2012: Your form does represent all but a finite set of primes, indeed all but a finite set of squarefree numbers not divisible by 11. If you take your original $w=0,$ the result is the ternary form $4 x^2 + 2 x y + 3 y^2 + 3 z^2,$ which reduces in the oneline BrandtIntrau order as $\langle 3,3,4,0,2,0 \rangle.$ The full genus and some relevant information:
=====Discriminant 132 ==Genus Size== 3
132: 1 3 11 0 0 0
132: 2 3 7 2 0 2
132: 3 3 4 0 2 0
size 3
The smallest numbers NOT represented by full genus
22 66 77 88 110 143 187 198 209 231
264 308 319 330 352 385 429 440 451 473
506 550 561 572 594 627 671 682 693 715
748 792 803 814 836 869 913 924 935 957
990
Disc: 132
==================================

132: 3 3 4 0 2 0
misses, compared with full genus
1 2 10 11 13
14 29 38 41 46
61 65 121 122 154
166 242 269 286 290
346 374 409 517 601
902 946 965
As you can see, the genus (collectively) represents all numbers except $11^{2k+1} \; \cdot \{11m + 2,6,7,8,10 \}$ for $k,m \geq 0.$
Now, the fundamental result of Duke and SchulzePillot says that any given positive ternary represents all sufficiently large numbers that are primitively represented by at least one form in the same spinor genus. In this case, the spinor genus and genus coincide. Some form in the genus represents any number not divisible by 11, and if the number is squarefree, the representation is primitive. The sad news is that the implied constant in "sufficiently large" is unknown and unknowable, so the bound is sometimes called "ineffective."
I do not see the D_SP paper on the arXiv, it rather precedes that, anyway W. Duke and R. SchulzePillot, Representations of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids, Inventiones Mathematicae vol. 99 (1990) pages 4957. We are using the Corollary to Theorem 3 on page 56.
EDIT TOOOO: I found a nice item about Tartakowsky's Theorem, so we are in good shape. In fact, your form represents all sufficiently large numbers not divisible by 121, and we do not otherwise need to worry about square factors. See On Explicit Versions of Tartakovski's Theorem by Rainer SchulzePillot, preprint available at PREPRINTS

$\begingroup$ @Will: Once you know that all squarefree numbers not divisible by 11 are represented, you can make the result effective as we are dealing with four variables (cf. my response below). Also, can you explain the meaning of the entries in your table and what software was used? $\endgroup$ Feb 19, 2012 at 20:11

$\begingroup$ @GH, I found a preprint about Tartakowsky's theorem and effective bounds by SchulzePillot that I think you will enjoy, I just put in his link. The software is mine. The order is $\langle a,b,c,r,s,t \rangle$ meaning $f(x,y,z) = a x^2 + b y^2 + c z^2 + r y z + s z x + t x y,$ as in J. Larry Lehman, Math. Comp. (1992) pages 399417. Let's see, one program (Magma) says that the genus consists of three classes. The theorem of Jones, quantitative in that of Siegel, says that every number locally represented by a form in the genus is integrally represented by at least one form in the genus. $\endgroup$ Feb 19, 2012 at 20:21

$\begingroup$ @GH more. The three forms/classes are thus $$x^2 + 3 y^2 + 11 z^2, \; \; 2 x^2 + 3 y^2 + 7 z^2 + 2 y z + 2 x y, \; \; 3 x^2 + 3 y^2 + 4 z^2 + 2 z x. $$ My C++ postprocess programs simply collect represented numbers up to the bound (probably 1000) and say what numbers are not represented, the "progressions." In this case, it is easy to prove the impression given by the computer printout. Finally, I have the machine show that $ 3 x^2 + 3 y^2 + 4 z^2 + 2 z x$ fails to integrally represent $1,2,10,11,13,14,\ldots$ although these numbers are integrally represented by another form in the genus. $\endgroup$ Feb 19, 2012 at 20:29

$\begingroup$ @GH, Alexander Schiemann psoted his original tables of positive ternary forms on the catalog of lattices, for this genus see math.rwthaachen.de/~Gabriele.Nebe/LATTICES/Brandt_1.html and search for the text $$ $$ B.I.discr = 132 = 1 *2^2 *3 *11 $\endgroup$ Feb 19, 2012 at 21:20

$\begingroup$ @Will Jagy: Thank you very much for your answer. I found a preprint about Tartakowsky's theorem and effective bounds by SchulzePillot. The effective bound of my form is $127540832401$ and it is enough to check uo to the number. $\endgroup$ Feb 20, 2012 at 9:34
As GH suggests, here the relevant Eisenstein and cusp spaces are small enough that everything can be done explicitly. It's even a bit better than the dimensions $5+4$ suggest, because our quadratic form is isodual, which puts its theta series in an eigenspace for the AtkinLehner involution $w_{44}$. The resulting formula is particularly nice for $n$ prime, and immediately shows that every prime other than $2$ and $11$ is represented, and indeed the number of representations is proportional to the number of points modulo the prime of an elliptic curve of conductor $11$.
Namely: let $$ E_2(q) = 1  24 \sum_{n=1}^\infty \frac{nq^n}{1q^n}; $$ this is not a modular form, but for every factor $d44$ the combination $$ \varepsilon^{(d)}_2(q) := d \cdot E_2(q^d)  \frac{44}{d} E_2(q^{44/d}) $$ is a weight2 form for $\Gamma_0(44)$. Let $$ \phi(q) = q \prod_{n=1}^\infty \bigl( (1q^n)(1q^{11n}) \bigr)^2 = q  2 q^2  q^3 + 2 q^4 + q^5 + 2 q^6  2 q^7 \cdots $$ be the unique eigencuspform for $\Gamma_0(11)$, associated to the elliptic curve $E: y^2+y=x^3x^2$ of discriminant $11$. Then the theta function $\sum_{n=0}^\infty r(n) q^n$ is $$ \frac {\varepsilon^{(1)}_2(q)  \varepsilon^{(2)}_2(q) + \varepsilon^{(4)}_2(q)} {30}  \frac45\bigl(\phi(q)+3\phi(q^2)+4\phi(q^4)\bigr). $$ The coefficients are obtained by matching $q$expansions to $O(q^{125})$, which is more than enough to prove that two weight$2$ forms on $\Gamma_0(44)$ coincide. In particular, for the number of representations of a prime $p$ other than $2$ and $11$ we have $$ r(p) = \frac45 (p + 1  a_p) $$ which is positive because $p+1  a_p$ is the number of points on $E \bmod p$ (which is indeed divisible by $5$ because $E$ has a rational $5$torsion point $x=y=0$).

$\begingroup$ This is very cool, thanks. Two simpleminded questions: 1. What does isodual mean? 2. Do you use any software to match coefficients up to $O(q^{125})$? $\endgroup$ Feb 20, 2012 at 19:55

$\begingroup$ Noam, a better question than the ones I just deleted. My experiments suggest that with a positive binary $f(x,y),$ the set of numbers integrally represented by the quaternary $$ f(x,y) + f(z,w) $$ is completely multiplicative, if some $m$ and $n$ are represented then so is $mn.$ I know this for principal forms. I thought the class number for the binary was important, but perhaps not. Something about quaternions... $\endgroup$ Feb 20, 2012 at 21:05

1$\begingroup$ @Keerthi M P & GH: thanks! @GH: 1) isodual = L is isomorphic to a scaling of its dual lattice. 2) I wrote a few lines of gp; I can post them later. @Will J: The general case can't be that much harder. For example if $n$ is squarefree and coprime to $22$ then we get $4(\prod_{pn}(p+1)  \prod_{pn} a_p)/5$ which is positive because $a_p < p+1$ for each $p$. [In the next edit I'll also say explicitly that $a_p$ is the $q^p$ coefficient of $\phi$...] $\endgroup$ Feb 20, 2012 at 21:07

1$\begingroup$ @Will Jagy: Not true in general. Try $f(x,y)=3x^2+7y^2$, $m=3$, $n=7$. $\endgroup$ Feb 20, 2012 at 21:31

1$\begingroup$ Well it's not hard to check from the formula that $11^n$ and $2 \cdot 11^n$ are the only missing values, which does imply that the form is completely multiplicative, but I don't expect a direct proof. $\endgroup$ Feb 21, 2012 at 0:50
There is a standard way to decide this question using modular forms or the Kloosterman refinement of the circle method (although the computational details might be tiresome).
If $r(n)$ denotes the number of representations over the integers, then $f(z)=\sum_{n=0}^\infty r(n)e(nz)$ is a modular form in $M_2(\Gamma_0(44))$, see Corollary 4.9.5 in Miyake: Modular Forms. We can write $f(z)$ as a linear combination of Eisenstein series and cuspidal Hecke eigenforms (including oldforms). Correspondingly, $r(n)$ decomposes uniquely as $r_\text{gen}(n)+r_\text{cusp}(n)$. Let us assume that $n$ is squarefree and coprime with $22$. Then $r_\text{gen}(n)$ is supported on a union of arithmetic progressions mod $968$ and it is either zero or $\gg n^{1\epsilon}$. In contrast, $r_\text{cusp}(n)\ll n^{1/2+\epsilon}$. In both estimates the implied constant depends only on $\epsilon>0$ and is effective, hence we can decide which $n$'s (squarefree and coprime with $22$) are represented. I should add that $r_\text{gen}(n)=0$ implies $r(n)=0$, hence in fact $r(n)\gg n^{1\epsilon}$ when $r(n)>0$.
P.S. Perhaps the computational details are not so bad: the Eisenstein subspace of $M_2(\Gamma_0(44))$ has dimension $5$, while the cuspidal subspace has dimension $4$.

$\begingroup$ @GH: Thank you very much for your answer. Is it enough to take the group $\Gamma_0(44)$? $\endgroup$ Feb 20, 2012 at 9:35

$\begingroup$ @unknown: You are absolutely right, I have updated my response. $\endgroup$ Feb 20, 2012 at 16:25