# What numbers are integrally represented by $4 x^2 + 2 x y + 7 y^2 - z^3$

This is related to my first MO question and Kevin Buzzard's conjecture at Integers not represented by $2 x^2 + x y + 3 y^2 + z^3 - z$

In December 2010 my question appeared in the M.A.A. Monthly, show that $4 x^2 + 2 x y + 7 y^2 - z^3 \neq \pm 2 m^3, \; \pm 32 m^3$ when $m$ has certain prime factorizations. The answers were due by April 2011 so I feel willing to mention it, although the answers have not appeared yet, I just got the August-September issue. Sigh.

A couple of days ago I thought I might check for identities, and found several good ones, showing that all odd numbers are represented for example. I believe there is no chance of completing this problem by identities owing to the non-represented numbers. So, that is the question, can anyone prove that $4 x^2 + 2 x y + 7 y^2 - z^3$ integrally represents everything else?

For verisimilitude, we have:

$$\begin{array}{cc} x = 4 n^3 - 18 n^2 + 3 n - 21, & y = -16 n^3 - 18 n + 1, \\\ z = 12 n^2 + 12, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+1. \end{array}$$

$$\begin{array}{cc} x = 4 n^3 - 42 n^2 - 73 n - 359, & y = -16 n^3 - 48 n^2 - 146n - 111, \\\ z = 12 n^2 + 24n+ 88, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n-3. \end{array}$$

$$\begin{array}{cc} x = 4 n^3 + 42 n^2 - 65 n + 417, & y = -16 n^3 + 48 n^2 - 166n + 137, \\\ z = 12 n^2 - 24n+ 98, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+5. \end{array}$$

$$\begin{array}{cc} x = 16 n^3 - 12 n^2 + 23 n + 6, & y = 8 n^3 - 24 n^2 + 28n - 27, \\\ z = 12 n^2 - 12 n+ 17, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n+10. \end{array}$$

$$\begin{array}{cc} x = 16 n^3 - 12 n^2 + 3 n + 1, & y = 8 n^3 - 24 n^2 + 18n - 7, \\\ z = 12 n^2 - 12 n+ 7, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n-10. \end{array}$$

$$\begin{array}{cc} x = 72 n^3 + 60 n^2 + 13 n, & y = -72 n^3 - 24 n^2 + 2 n + 1, \\\ z = 36 n^2 + 12 n+ 1, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n + 6. \end{array}$$

$$\begin{array}{cc} x = 4 n^3 + 36 n^2 + 18 n + 135, & y = -16 n^3 - 60 n + 4, \\\ z = 12 n^2 + 42, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 24n + 4. \end{array}$$

$$\begin{array}{cc} x = 9 n^3 - 30 n^2 + 29 n - 16, & y = -9 n^3 + 12 n^2 - 8 n + 2, \\\ z = 9 n^2 -12 n + 10, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 36n - 12. \end{array}$$

$$\begin{array}{cc} x = 16 n^3 - 12 n^2 + 33 n + 7, & y = 8 n^3 - 24 n^2 + 30 n - 37, \\\ z = 12 n^2 -12 n + 21, & 4 x^2 + 2 x y + 7 y^2 - z^3 = 162 n. \end{array}$$

Furthermore, if we have a prime $q = 4 u^2 + 2 u v + 7 v^2,$ the fact that $h(-108) = 3$ and $2^2 + 27 \cdot 1^2 = 31$ shows that $4 x^2 + 2 x y + 7 y^2$ represents $q^3, \; 31 q^3, \; 25 q^3.$ As a result $q = 4 u^2 + 2 u v + 7 v^2 - z^3$ represents $2 q^3 = q^3 + q^3, \; 32 q^3 = 31 q^3 + q^3, \; -2 q^3 = 25 q^3 - 27 q^3.$ I'm not sure how to do $-32 q^3.$

P.S. Not that it really increases the difficulty, but representing $\pm 2 q^3, \pm 32 q^3$ is not actually enough... if we can represent some $n,$ for any $k$ we know we can also represent $n k^6,$ but not necessarily $n k^3.$ I'm just saying.

P.P.S. Komputer Kalkulation:

 Targets between  -1,000,000  and  1,000,000
that appear to have no integer expression as
4 x^2 + 2 x y + 7 y^2 + z^3  :

-953312 =  -1 * 2^5 * 31^3
-780448 =  -1 * 2^5 * 29^3
-715822 =  -1 * 2 * 71^3
-500000 =  -1 * 2^5 * 5^6
-410758 =  -1 * 2 * 59^3
-389344 =  -1 * 2^5 * 23^3
-332750 =  -1 * 2 * 5^3 * 11^3
-297754 =  -1 * 2 * 53^3
-207646 =  -1 * 2 * 47^3
-159014 =  -1 * 2 * 43^3
-157216 =  -1 * 2^5 * 17^3
-137842 =  -1 * 2 * 41^3
-59582 =  -1 * 2 * 31^3
-48778 =  -1 * 2 * 29^3
-42592 =  -1 * 2^5 * 11^3
-31250 =  -1 * 2 * 5^6
-24334 =  -1 * 2 * 23^3
-9826 =  -1 * 2 * 17^3
-4000 =  -1 * 2^5 * 5^3
-2662 =  -1 * 2 * 11^3
-250 =  -1 * 2 * 5^3
-32 =  -1 * 2^5
-2 =  -1 * 2
2 = 2
32 = 2^5
250 = 2 * 5^3
2662 = 2 * 11^3
4000 = 2^5 * 5^3
9826 = 2 * 17^3
24334 = 2 * 23^3
31250 = 2 * 5^6
42592 = 2^5 * 11^3
48778 = 2 * 29^3
59582 = 2 * 31^3
137842 = 2 * 41^3
157216 = 2^5 * 17^3
159014 = 2 * 43^3
207646 = 2 * 47^3
297754 = 2 * 53^3
332750 = 2 * 5^3 * 11^3
389344 = 2^5 * 23^3
410758 = 2 * 59^3
500000 = 2^5 * 5^6
715822 = 2 * 71^3
780448 = 2^5 * 29^3
953312 = 2^5 * 31^3

phoebus:~/Cplusplus>


Monday, August 20: A student of Kevin Buzzard, in what would be a Master's thesis in the U.S., proved that for any integers $A,B,$ both the inhomogeneous polynomials $$x^2 + x y + 6 y^2 + z^3 + A z^2 + B z$$ and $$x^2 + x y + 8 y^2 + z^3 + A z^2 + B z$$ are universal, they integrally represent all integers. He also did a fixed one, $$2x^2 + x y + 2 y^2 + z^3 + z.$$ So the hard case really is these non-universal ones.

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+1 for ask-kevin-buzzard – Steven Gubkin Aug 17 '12 at 3:38
Am I the only one that finds this use of tags strange? – quid Aug 17 '12 at 20:58
@quid, it started by request, of a sort, at the earliest of these: mathoverflow.net/questions/tagged/ask-johnson where all involved intended it as humorous. I tried this at MSE math.stackexchange.com/questions/93600/… where some local lout deleted it without consulting me. I like your way better. I would like to give it some time before deleting them, though, perhaps a day, this question and related ones have been on my mind for several years. Kevin and Matt are now both intermittent MO participants. – Will Jagy Aug 17 '12 at 21:53
@Will Jagy: thanks for the reply. Let me say it like this: if such tagging stays an exception I can live with it without problem. If it should however become some general trend, I'd find it anoying. One other thing: you do not have a top-level arXiv tag, like nt.number-theory, and no more space for it. I do not know how K.B. and M.E. read MO. But, if I'd not be frequently around, I'd just look at my preferred top-level tags. So, it could even be that on a practical level a standard NT-tag would help more than the personalized ones. In any case, I hope you get a good answer. Good luck! – quid Aug 18 '12 at 0:23
Are there known obstructions a prime to be both $z^3 \pm 2$ and $4 u^2 + 2 u v + 7 v^2$ ? – joro Aug 21 '12 at 15:16