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Hilbert 10th problem, asking for algorithm for determining whether a polynomial Diopantine equation has an integer solution, is undecidable in general, but decidable or open in some restricted families. If we classify equations by degree, the problem is trivial for linear equations, decidable for quadratic equations by the Theorem of Grunewald and Segal, but is open for cubic equations.

It is interesting to investigate what are the ``smallest'' difficult cubic equations, in the spirit of this previous question for general equations. For a polynomial $P$ consisting of $k$ monomials of degrees $d_1,\dots,d_k$ and integer coefficients $a_1,\dots,a_k$, let $H(P)=\sum_{i=1}^k |a_i|2^{d_i}$. If we order all equations by $H$, then the smallest cubic equation I currently cannot solve is the equation $$ 3-y+x^2 y+y^2+x y z-2 z^2 = 0, $$ of size $H=33$.

This equation after a linear substitution reduces to $$ 2 x^2 - y x z + 2 z^2 = y^2 - 9 y + 23, $$ which is nice because it is symmetric in $x,z$. Vieta jumping seems to be difficult to apply because of coefficients $2$ near $x^2$ and $z^2$.

The question is whether this equation has any integer solutions.

Bogdan Grechuk
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  • Erdős–Straus conjecture (https://en.m.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture) involves cubic equations with parameter. For each specific value of parameter you can solve equation in finite time, but whether it has solutions for all values of parameter is still an open problem. While this doesn't answer the question, it may indicate that cubic equations can be hard. – Somnium Apr 11 '22 at 22:30
  • The only way to solve such equations. This is to make such transformations that would lead to the Pell equation. Then, after some equivalent Pell equation, express its solutions. Sometimes a record can be expressed in terms of several Pell equations. And the formula will be very cumbersome. The answer will be in the solution of the Pell equations. At the same time, its coefficients will be solutions of some Pell equation. If it is possible to break the connection between these two equations and make them independent, then the formula can be written. – individ Apr 12 '22 at 05:32
  • On the other hand, I do not understand and do not see the logic in the fact that you ask similar questions to each other. From the answers you choose the solutions that you like. Or the absence of equations with solutions. And what you don't like, you ignore. Nature doesn't care if you like its laws or not. And if you ignore everything that you don't like, it will be impossible to understand how the phenomenon functions. – individ Apr 12 '22 at 05:39
  • What is the smallest open qubic equation in $\ge 4$ variables? – Denis Shatrov Aug 23 '23 at 11:59
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    In response to your question, I run my program up to $H\leq 39$ for equations in $n\geq 4$ variables, and the only non-trivial equations it finds are $1+t^2+t x^2+y^2+x^2 y z+z^2=0$ of size $H=37$ and $3+t^2+x^2+y^2+y z+x^2 y z+z^2=0$ of size $H=39$. I did not try to solve them by hand yet, so I am not sure how interesting they are. Both equations are quartic, hence there are no open cubic equations in this many variables in this range. Deeper search would require some time, because the number of equations increases exponentially with $H$. – Bogdan Grechuk Aug 24 '23 at 11:14
  • Thank you. This is interesting. I thought that the solvability problem for qubic equations could be easier if only equations in $\ge 4$ variables are considered. Today I found that the problem of representation of every integer as the sum of 4 cubes is still open. But it is known that every integer is the sum of 5 cubes. Maybe, there is an algorithm for solving a cubic equation if the number of variables is big enough. – Denis Shatrov Aug 24 '23 at 12:41

1 Answers1

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Update. I think, now the solution to this beautiful equation is complete.

The equation $$2 x^2 - y x z + 2 z^2 = y^2 - 9 y + 23$$ is equivalent to $$X^2 - f_1Y^2 = 8f_2$$ where $X = 4x - yz$, $Y = z$, $f_1 = y^2 - 16$, $f_2 = y^2 - 9y + 23$.

A few simple statements. From the original equation it is not hard to see that $x$, $y$, $z$ are odd. Let $d = \gcd(f_1, f_2)$. Then $d \mid 75$. If $d \in \{5, 15\}$, then $\gcd(f_1, 25) = \gcd(f_2, 25) = 5$. If $3 \mid d$, then $f_1 \equiv 6 \pmod{9}$, $9 \mid f_2$ (in other cases the equation has no solutions modulo 9).

It is always possible to prove the absence of solutions by considering an appropriate Jacobi symbol.

If $d \in \{1, 3, 5, 15\}$, then $\left(\frac{8f_2}{f_1/d}\right) = -1$

If $d \in \{25, 75\}$, then $5 \mid X$, $\left(\frac{X}{5}\right)^2 - \frac{f_1}{25}Y^2 = \frac{8f_2}{25}$ and $\left(\frac{8f_2/25}{f_1/d}\right) = -1$

The cruicial observation is that the numbers in these Jacobi symbols are coprime, as follows from previous statements. I will consider the cases $d = 1$ and $d = 3$. In the remaining cases, the proof is similar.

Case $d = 1$. $$\left(\frac{8(y^2 - 9y + 23)}{y^2 - 16}\right) = \left(\frac{y^2 - 9y + 23}{y - 4}\right) \left(\frac{y^2 - 9y + 23}{y + 4}\right) = \left(\frac{3}{y - 4}\right)\left(\frac{3}{y + 4}\right) = -1$$ since $y - 4 $ and $y + 4 $ have different remainders modulo 3 and same remainders modulo 4.

Case $d = 3$.

$y^2 - 16 = f_1 \equiv 6 \pmod{9}$. Therefore $y \equiv \pm 2 \pmod{9}$.

$$\left(\frac{8(y^2 - 9y + 23)}{(y^2 - 16)/3}\right) = -\left(\frac{y^2 - 9y + 23}{(y \pm 4)/3}\right) \left(\frac{y^2 - 9y + 23}{y \mp 4}\right) = -\left(\frac{3}{(y \pm 4) /3}\right)\left(\frac{3}{y \mp 4}\right) = -1$$ since $(y \pm 4)/3$ and $y \mp 4$ have different remainders modulo 3 and different remainders modulo 4.

Denis Shatrov
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