On the equation $7x^3 + 2y^3 = 3z^2 + 1$

Question

The question is whether there exist integers $x,y,z$ such that $$ 7x^3+2y^3=3z^2+1. $$ After a similar equation On the equation $9x^3+y^3=z^2+3$ has been solved, this is one of the nicest cubic equations for which it is not known whether integer solutions exist.

In fact, such equations either have many not-too-large solutions or no integer solutions at all, hence I am quite confident that there are no integer solutions. For such equations, this can often be proved by cubic reciprocity. However, the fact that $3z^2 + 1$ may be divisible by an arbitrary large power of $7$ makes cubic reciprocity more difficult to apply.

Answer 1

This is not a solution, but a promising approach to the problem. I tried to apply the method I used to solve the equation $$9x^3 + y^3 = z^2 + 3$$

The following conjecture seems to be true. Let $3z^2 + 1 = 7^a \cdot b$ where $b$ is not divisible by 7. If the following conditions are satisfied, then $b$ has prime divisor $p$ such that 28 is a cubic nonresidue modulo $p$ and $\nu_p(b)$ is not divisible by 3.

1. $z$ is even and not divisible by 3
2. $a \equiv 1 \pmod{3}$ or $a \equiv 0 \pmod{3}$
3. If $a \equiv 0 \pmod{3}$, then $b \equiv \pm 2 \pmod{7}$. If $a \equiv 1 \pmod{3}$, then $b \equiv \pm 1 \pmod{7}$.

If we have triple $(x, y, z)$ such that $$7x^3 + 2y^3 = 3z^2 + 1$$ then, by considering this equation modulo 8, 9 and modulo powers of 7, we see that $3z^2 + 1$ satisfies all conditions of the conjecture. Hence $3z^2 + 1$ has prime divisor $p$ such that 28 is a cubic nonresidue modulo $p$ and $\nu_p(b)$ is not divisible by 3. This is a contradiction, because from the equation it follows that $3 \mid \nu_p(b)$.

The conjecture is true for $|z| < 10^9$. The problem is to prove this conjecture. Let $$b = c^2 + 3d^2 = (c + d + 2d\omega)(c - d - 2d\omega)$$ It is possible to prove that $\left(\frac{4}{1 + z + 2\omega}\right)_3 = \omega^2$. Then we can calculate $\left(\frac{4}{c + d + 2d\omega}\right)_3$. But I was not able to find the value of $\left(\frac{7}{c + d + 2d\omega}\right)_3$.