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The question is whether there exist integers $x,y,z$ such that $$ 9x^3+y^3=z^2+3. $$ This is one of the nicest (if not the nicest one!) cubic equations for which I do not know whether integer solutions exist.

Heuristic argument predicts a lot of solutions, but the search returned no integer solutions up to $|z|\leq 50,000$. It seems that there must be a reason for this. ${}{}{}{}$

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    $\begingroup$ Possibly useful note for extending the search range: Mod 9 shows that $z \not \equiv 0\pmod{3}$, and one can use that to then show that $y \equiv 1 \pmod{3}$. $\endgroup$
    – JoshuaZ
    Commented Feb 18 at 20:07
  • $\begingroup$ well, what about $9 x^3 + y^3 = z^2 + 3 w^2 $ $\endgroup$
    – Will Jagy
    Commented Feb 18 at 20:29
  • $\begingroup$ Of course $9x^3+y^3=z^2+3w^2$ is solvable. The simplest solution is (0,0,0,0). Another example is (1,3,3,3). $\endgroup$
    – Bogdan Grechuk
    Commented Feb 18 at 20:34
  • $\begingroup$ Good. Next restriction, with $\gcd(z,w) = 1.$ If this is difficult or impossible it may explain some things $\endgroup$
    – Will Jagy
    Commented Feb 18 at 20:36
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    $\begingroup$ ran a little program, lots of those... $\endgroup$
    – Will Jagy
    Commented Feb 18 at 20:51

1 Answer 1

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This equation is unsolvable. Modulo 9 analysis shows that $z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$. Let $p$ be a prime divisor of $z^2 + 3$ for which 3 is not a cubic residue. From $$z^2 + 3 = 1/3(3y^3 - (-3x)^3)$$ it follows that $3 \mid \nu_p(z^2 + 3)$. But we will show that $z^2 + 3$ ($z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$) always has prime divisor $q$ for which 3 is a cubic nonresidue and $\nu_q(z^2 + 3)$ is not divisible by 3.

The proof requires the ring $\mathbb{Z}[\omega]$, $\omega = \frac{-1 + i\sqrt{3}}{2}$. $$z^2 + 3 = (z + 1 + 2\omega)(z - 1 - 2\omega)$$ We can calculate cubic character using the cubic reciprocity law. Let $z \equiv 1 \pmod{3}$, replace $z$ with $-z$ otherwise. \begin{equation*} \begin{split} \left(\frac{3}{z + 1 + 2\omega}\right)_3 = \left(\frac{3}{\omega(z + 1 + 2\omega)}\right)_3 = \\ = \left(\frac{3}{-2 + (z - 1)\omega}\right)_3 = \omega^{1/3(z - 1)} \ne 1 \\ \end{split} \end{equation*} Hence, $z + 1 + 2\omega$ has prime divisor $\pi$ such that 3 is a cubic nonresidue modulo $\pi$ and $\nu_{\pi} (z + 1 + 2\omega)$ is not divisible by 3. The same applies to complex conjugate $\overline{\pi} \mid z - 1 - 2\omega$. Thus, we can take $q = \pi \overline{\pi} \in \mathbb{Z} $.

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  • $\begingroup$ Can you say more about why $3$ being a cubic nonresidue mod $p$ implies that $3\mid \nu_p(z^2+3)$? I don't follow that step. $\endgroup$
    – Timothy Chow
    Commented Feb 22 at 19:31
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    $\begingroup$ @TimothyChow If $y$ is not divisible by $p$, then $3 \equiv (\frac{-3x}{y})^3 \pmod{p}$ which is impossible. If $y \mid p$, then $x \mid p$. $9(x/p)^3 + (y/p)^3 = (z^2 + 3)/p^3$. We can continue this process until we get $9(x/p^k)^3 + (y/p^k)^3 = (z^2 + 3)/(p^{3k}) $ where $(z^2 + 3)/(p^{3k})$ is not divisible by $p$. Hence, $\nu_p(z^2 + 3) = 3k$. $\endgroup$
    – Denis Shatrov
    Commented Feb 22 at 21:15
  • $\begingroup$ A typo in second sentence: If $p \mid y$ then $p \mid x$. $\endgroup$
    – Denis Shatrov
    Commented Feb 22 at 21:27

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