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EDIT: a selection of material relevant to this problem is available at INHOMOGENEOUS

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 appeared in the December 2012 issue. Exactly one other person got it right, Robin Chapman.

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> 

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.

NOTE: if you prefer, essentially the same question for $x^2 + 27 y^2 - 7 z^3.$

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    @WillJagy The tag (complex-multiplication) is now [discussed on meta](http://meta.math.stackexchange.com/questions/22348/tag-management-2016/22586#22586).2016-02-11

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