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Do there exist integers $x,y,z$ satisfying $$ (3x-1)y^2 + x z^2 = x^3-2 \quad ? $$

Hilbert's 10th Problem is unsolvable in general, but is still open for cubic equations: it is unknown whether there exists an algorithm that decides whether a cubic equation has an integer solution.

If we order the equations by ``length'' (equal to sum of degrees of monomials plus sum of logarithms of absolute values of the coefficients), then the current shortest open cubic equation is $7x^3+2y^3=3z^2+1$, see On the equation $7x^3 + 2y^3 = 3z^2 + 1$ . The equation in the title is one of the next-shortest ones.

For each fixed $x$, the equation is quadratic in $(y,z)$, and normally such equations either have a solution with reasonable small $x$, or have no solutions because of something like quadratic residues argument. But this equation has no solutions for $y$ up to $500$ millions (updated: $3$ billions), and resisted all my attempts so far to prove its unsolubility. Try it!

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    $\begingroup$ This affine varietiy has rational points, like $(4/3,1/3,1/6)$ which, presumably, makes Brauer-Manin obstruction "stuff" harder to use. $\endgroup$ Commented Mar 10 at 9:48
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    $\begingroup$ For what it worth, $x$ can not be odd: we have $y^2\equiv 2\pmod x$, thus all prime divisors of $x$ are $\pm 1 \pmod 8$, thus so is $x$, but both cases are impossible modulo 8 $\endgroup$ Commented Mar 10 at 10:44
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    $\begingroup$ Yes, moreover, modulo $16$ analysis return only two even values for $x$, $2$ and $6$. Because all prime divisors of $x/2$ are $\pm 1$ mod $8$, $6$ is impossible. Further, modulo $9$ analysis shows that $x$ is $1$ mod $3$. Hence, $x$ is $34$ mod $48$. $\endgroup$ Commented Mar 10 at 13:50
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    $\begingroup$ Since $3x-1$ and $x$ have to have the same sign, the equation is positive definite in $y,z$ and so $y$ and $z$ both must be $O(x)$. The heuristic probability that a solution exists for fixed $x$ is going to be $O(1/x)$, and then unless the constant in the big $O$ is very large, summing only over $x$ congruent to $34$ mod $48$ is going to give a prediction for the number of solutions with $x<X $ that is a small constant times $\log X$. So maybe the search just isn't big enough. $\endgroup$ Commented Mar 10 at 18:38
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    $\begingroup$ Now checked for |y| up to three billions ($3\cdot 10^9), still no solutions. I will stop search here, at least temporary. $\endgroup$ Commented Mar 14 at 18:30

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