The new shortest open cubic equations

Question

Section 8.3.2 of recent book [1] studies the following problem. Define the length of a Polynomial Diophantine equation as the sum of degrees of monomials plus sum of base 2 logarithms of the coefficients, order the equations by length, and investigate the existence of integer solutions (you do not need to find all solutions). The shortest cubic equation left open in the book is $$ 1 + 3 x^3 + x y^2 + 6 y z^2 = 0 \quad\quad\quad (1) $$ The question is whether there exist integers $x,y,z$ satisfying this equation. With $t=xy$, this reduces to the solvability of $x+3x^4+t^2+6tz^2=0$ such that $x$ is a divisor of $t$.

Update 20.09.2024: Dmitry Ezhov now found some solutions to (1) starting from $(x,y,z) =(-1017461719,95574914,2350866170)$, see comments. So, this equation is now resolved. The next-shortest open cubic equation in [1] is
$$ y^2+10xyz+x^3-x-2=0. \quad\quad\quad (2) $$ Modulo $20$ analysis shows that $x=2$ mod $20$. After replacing $y$ with $-y$ if necessary, we may also assume that $y=4$ mod $10$.

Update 07.10.2024: Dmitry Ezhov now found a solution $$ (23499130751021842,252697047241468990409432149765008357514,-1075346360334969622883) $$ to (2). A solution of this size would be unfeasible to find by direct search, but nice ascent method for the related equation $y^2+xyz+x^3-x-2=0$ has been used, see comments.

The next-shortest open cubic equation in [1] is
$$ y^2+7xyz+3x^3-2=0. \quad\quad\quad (3) $$ Modulo $4$ and $7$ analysis shows that $x=7$ mod $14$. After replacing $y$ with $-y$ if necessary, we may also assume that $y=3$ mod $14$. Does this equation have any integer solutions? Can a solution be found by first developing ascent formulas for the solutions to $y^2+xyz+3x^3-2=0$?

Update 10.10.2024: Equation (3) has a solution, see the answer. Now the shortest open cubic is the equation $$ 7x^3+2y^3=3z^2+1 \quad\quad (4) $$ which I have already asked to solve, see On the equation $7x^3 + 2y^3 = 3z^2 + 1$

[1] Bogdan Grechuk, Polynomial Diophantine equations. A systematic approach, Springer, 2024

Answer 1

Equation $$ y^2+7xyz+3x^3-2=0 $$ is solvable in integers. Take, for example, $$ x = 47699434725285831080938680589289, $$ $$ y = 80298610335148427555, $$ $$ z = -12143437727264755796194424115079504569776282. $$

Verification is straithforward, but here is the method this solution has been found. Assume that $(x,y,z)$ is a solution with $y$ odd. Then, obviously, $$ y^2-2 \equiv 0 \,(\text{mod}\,\,x) \quad \text{and} \quad 3x^3-2 \equiv 0 \, (\text{mod}\,\,y). $$ Then $X=\frac{y^2-2}{x}$ is an odd integer. We have $$ 0 \equiv X^3(3x^3-2) = X^3\left(3\left(\frac{y^2-2}{X}\right)^3-2\right)=3(y^2-2)^3-2X^3 \equiv -24-2X^3 \, (\text{mod}\,\,y). $$ Because $y$ is odd, we conclude that $$ y^2-2 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,y). $$ Then $Y=\frac{X^3+12}{y}$ is an odd integer. Then $$ 0 \equiv Y^2(y^2-2) = Y^2\left(\left(\frac{X^3+12}{Y}\right)^2-2\right)=(X^3+12)^2-2Y^2 \equiv 144-2Y^2 \, (\text{mod}\,\,X). $$ Because $X$ is odd, we conclude that $$ Y^2-72 \equiv 0 \,(\text{mod}\,\,X) \quad \text{and} \quad X^3+12 \equiv 0 \, (\text{mod}\,\,Y). $$ Now let us just try odd integers $Y=1,3,5,\dots$, try $X$ divisors of $Y^2-72$, and check whether $X^3+12$ is divisible by $Y$. If it is, compute $y=\frac{X^3+12}{Y}$, then $x=\frac{y^2-2}{X}$. Then ratio $\frac{y^2+3x^3-2}{xy}$ must be an integer, and it is left to check whether this integer is divisible by $7$. For $Y$ as small as $30761$ this works: divisor $X=135177007$ returns the solution presented above.