Conjecture: $N=x_1^2+x_2^2-y^3$ has integer solutions for all $N$, with $x_1>0$,$x_2>0$ and $y>0$

Question

Background.

I constructed some code for Algorithm for diophantine equation and decided to reuse it to investigate $N=x_1^2+x_2^2+z^3$ with $z$ integer. Negative values for $z$ seemed to produce plentiful small $N$ values, so I elected to concentrate my efforts here, and changed $z$ to $-y$.

Question.

Can anyone prove or disprove this conjecture, or help me find a method to do this, please? I would also appreciate any useful background information.

My efforts.

Under nine minutes brute-force to find a solution for each $N=-10^6$ to $10^6$.

I’ve searched around on the net to find similar solutions. Perhaps a method based on this link http://www.dms.umontreal.ca/~mlalin/Lagrange.pdf would do?

Examples. $$0=2^2+11^2-5^3$$ $$11=6^2+10^2-5^3$$ $$-3=5^2+6^2-4^3$$ $$999999=40^2+1718^2-125^3$$ $$-999999=8^2+1365^2-142^3$$

Answer 1

This was a problem posed by Noam Elkies and Irving Kaplansky in the January 1995 M.A.A. Monthly. They did not ask that your $y$ be positive, but that is how it comes out anyway, with a very few exceptions for the variable $x$ below being, for example $0,1,2.$ These can then be fixed by a quick search.

Later, Kap and I published a little paper where we pointed out that $x^2 + y^2 - z^9$ fails to represent infinitely many numbers The closest looking thing to your original that fails to represent infinitely many numbers would be $x^2 + 27 y^2 - 7 z^3.$

Here is a solution by Andrew Adler, in the June-July 1997 issue: $$ 2x+1 = \left( x^3 - 3 x^2 + x \right)^2 + \left(x^2 - x - 1 \right)^2 -\left( x^2 - 2 x\right)^3 $$ $$ 4x+2 = \left(2 x^3 - 2 x^2 - x \right)^2 + \left(2 x^3 - 4 x^2 - x + 1 \right)^2 -\left( 2 x^2 - 2 x - 1 \right)^3 $$ $$ 8x+4 = \left(x^3 + x + 2 \right)^2 + \left(x^2 - 2x - 1 \right)^2 -\left(x^2 + 1 \right)^3 $$ $$ 16x+8 = \left( 2 x^3 - 8 x^2 + 4 x + 2 \right)^2 + \left(2 x^3 - 4 x^2 - 2 \right)^2 -\left(2 x^2 - 4 x \right)^3 $$ $$ 16x = \left(x^3 + 7 x - 2 \right)^2 + \left(x^2 + 2 x + 11 \right)^2 -\left( x^2 + 5 \right)^3 $$

Answer 2

This is just a follow-up. I mentioned that the polynomial $x^2 + 27 y^2 - 7 z^3$ fails to represent infinitely many numbers. I put a somewhat less intuitive version at What numbers are integrally represented by $4 x^2 + 2 x y + 7 y^2 - z^3$

Meanwhile, here are the numbers (between certain bounds) I think cannot be integrally represented by $x^2 + 27 y^2 - 7 z^3.$ The nice part is how they come in $\pm$ pairs, some $\pm 14 m^3$ and some $\pm 224 n^3.$ Meanwhile, this is all based on cubic reciprocity and the cubic character of two, which is in Ireland and Rosen for example. My own proof that things work out this way is long, long, long.

    NOT x^2 + 27 y^2 - 7 z^3
 -1113098 =  -1 * 2 * 7 * 43^3
 -1100512 =  -1 * 2^5 * 7 * 17^3
 -964894 =  -1 * 2 * 7 * 41^3
 -417074 =  -1 * 2 * 7 * 31^3
 -341446 =  -1 * 2 * 7 * 29^3
 -298144 =  -1 * 2^5 * 7 * 11^3
 -218750 =  -1 * 2 * 5^6 * 7
 -170338 =  -1 * 2 * 7 * 23^3
 -68782 =  -1 * 2 * 7 * 17^3
 -28000 =  -1 * 2^5 * 5^3 * 7
 -18634 =  -1 * 2 * 7 * 11^3
 -1750 =  -1 * 2 * 5^3 * 7
 -224 =  -1 * 2^5 * 7
 -14 =  -1 * 2 * 7
 14 = 2 * 7
 224 = 2^5 * 7
 1750 = 2 * 5^3 * 7
 18634 = 2 * 7 * 11^3
 28000 = 2^5 * 5^3 * 7
 68782 = 2 * 7 * 17^3
 170338 = 2 * 7 * 23^3
 218750 = 2 * 5^6 * 7
 298144 = 2^5 * 7 * 11^3
 341446 = 2 * 7 * 29^3
 417074 = 2 * 7 * 31^3
 964894 = 2 * 7 * 41^3
 1100512 = 2^5 * 7 * 17^3
 1113098 = 2 * 7 * 43^3
Thu Feb 15 12:59:32 PST 2018
    NOT x^2 + 27 y^2 - 7 z^3