This is related to my first MO question and Kevin Buzzard's conjecture at
http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-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.