What numbers are integrally represented by $4 x^2 + 2 x y + 7 y^2 - z^3$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:35:48Z http://mathoverflow.net/feeds/question/104889 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104889/what-numbers-are-integrally-represented-by-4-x2-2-x-y-7-y2-z3 What numbers are integrally represented by $4 x^2 + 2 x y + 7 y^2 - z^3$ Will Jagy 2012-08-17T03:31:48Z 2012-12-04T23:16:21Z <p>This is related to my first MO question and Kevin Buzzard's conjecture at <a href="http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z" rel="nofollow">http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z</a></p> <p>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.</p> <p>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 <strong>question</strong>, can anyone prove that $4 x^2 + 2 x y + 7 y^2 - z^3$ integrally represents everything else? </p> <p>For verisimilitude, we have:</p> <p>$$\begin{array}{cc} x = 4 n^3 - 18 n^2 + 3 n - 21, &amp; y = -16 n^3 - 18 n + 1, \\ z = 12 n^2 + 12, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+1.<br> \end{array}$$</p> <p>$$\begin{array}{cc} x = 4 n^3 - 42 n^2 - 73 n - 359, &amp; y = -16 n^3 - 48 n^2 - 146n - 111, \\ z = 12 n^2 + 24n+ 88, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n-3.<br> \end{array}$$ </p> <p>$$\begin{array}{cc} x = 4 n^3 + 42 n^2 - 65 n + 417, &amp; y = -16 n^3 + 48 n^2 - 166n + 137, \\ z = 12 n^2 - 24n+ 98, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 6n+5.<br> \end{array}$$ </p> <p>$$\begin{array}{cc} x = 16 n^3 - 12 n^2 + 23 n + 6, &amp; y = 8 n^3 - 24 n^2 + 28n - 27, \\ z = 12 n^2 - 12 n+ 17, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n+10.<br> \end{array}$$</p> <p>$$\begin{array}{cc} x = 16 n^3 - 12 n^2 + 3 n + 1, &amp; y = 8 n^3 - 24 n^2 + 18n - 7, \\ z = 12 n^2 - 12 n+ 7, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n-10.<br> \end{array}$$</p> <p>$$\begin{array}{cc} x = 72 n^3 + 60 n^2 + 13 n, &amp; y = -72 n^3 - 24 n^2 + 2 n + 1, \\ z = 36 n^2 + 12 n+ 1, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 18n + 6.<br> \end{array}$$</p> <p>$$\begin{array}{cc} x = 4 n^3 + 36 n^2 + 18 n + 135, &amp; y = -16 n^3 - 60 n + 4, \\ z = 12 n^2 + 42, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 24n + 4.<br> \end{array}$$</p> <p>$$\begin{array}{cc} x = 9 n^3 - 30 n^2 + 29 n - 16, &amp; y = -9 n^3 + 12 n^2 - 8 n + 2, \\ z = 9 n^2 -12 n + 10, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 36n - 12.<br> \end{array}$$</p> <p>$$\begin{array}{cc} x = 16 n^3 - 12 n^2 + 33 n + 7, &amp; y = 8 n^3 - 24 n^2 + 30 n - 37, \\ z = 12 n^2 -12 n + 21, &amp; 4 x^2 + 2 x y + 7 y^2 - z^3 = 162 n.<br> \end{array}$$</p> <p>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> <p>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>P.P.S. Komputer Kalkulation:</p> <pre><code> 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&gt; </code></pre> <p>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. </p>