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 Choi
    Feb 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 Jagy
    Feb 19, 2012 at 9:14
  • $\begingroup$ edited only to put the formula for the quadratic form in TeX. $\endgroup$ Feb 20, 2012 at 21:10

3 Answers 3


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 2-adic numbers but anisotropic in the 11-adic 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 one-line Brandt-Intrau 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
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 Schulze-Pillot 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_S-P paper on the arXiv, it rather precedes that, anyway W. Duke and R. Schulze-Pillot, Representations of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids, Inventiones Mathematicae vol. 99 (1990) pages 49-57. 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 Schulze-Pillot, preprint available at PREPRINTS

  • $\begingroup$ @Will: Once you know that all square-free 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? $\endgroup$
    – GH from MO
    Feb 19, 2012 at 20:11
  • $\begingroup$ @GH, I found a preprint about Tartakowsky's theorem and effective bounds by Schulze-Pillot 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 399-417. 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. $\endgroup$
    – Will Jagy
    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++ post-process 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. $\endgroup$
    – Will Jagy
    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.rwth-aachen.de/~Gabriele.Nebe/LATTICES/Brandt_1.html and search for the text $$ $$ B.-I.discr = -132 = -1 *2^2 *3 *11 $\endgroup$
    – Will Jagy
    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 Schulze-Pillot. The effective bound of my form is $127540832401$ and it is enough to check uo to the number. $\endgroup$
    – user21534
    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 Atkin-Lehner 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 $$ E_2(q) = 1 - 24 \sum_{n=1}^\infty \frac{nq^n}{1-q^n}; $$ this is not a modular form, but for every factor $d|44$ the combination $$ \varepsilon^{(d)}_2(q) := d \cdot E_2(q^d) - \frac{44}{d} E_2(q^{44/d}) $$ is a weight-2 form for $\Gamma_0(44)$. Let $$ \phi(q) = q \prod_{n=1}^\infty \bigl( (1-q^n)(1-q^{11n}) \bigr)^2 = q - 2 q^2 - q^3 + 2 q^4 + q^5 + 2 q^6 - 2 q^7 \cdots $$ be the unique eigen-cuspform for $\Gamma_0(11)$, associated to the elliptic curve $E: y^2+y=x^3-x^2$ 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 simple-minded questions: 1. What does isodual mean? 2. Do you use any software to match coefficients up to $O(q^{125})$? $\endgroup$
    – GH from MO
    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... $\endgroup$
    – Will Jagy
    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 $4(\prod_{p|n}(p+1) - \prod_{p|n} a_p)/5$ 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 square-free 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 (square-free 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 $\Gamma_0(44)$? $\endgroup$
    – user21534
    Feb 20, 2012 at 9:35
  • $\begingroup$ @unknown: You are absolutely right, I have updated my response. $\endgroup$
    – GH from MO
    Feb 20, 2012 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.