Do there exist integers $x,y,z$ such that $$ xy(x+y)=7z^2 + 1 ? $$ The motivation is simple. Together with Aubrey de Grey, we developed a computer program that incorporates all standard methods we know (Hasse principle, quadratic reciprocity, Vieta jumping, search for large solutions, etc.) to try to decide the solvability of Diophantine equations, and this equation is one of the nicest (if not the nicest) cubic equation that our program cannot solve.

1$\begingroup$ Is your program publicly available? $\endgroup$– Max AlekseyevCommented Apr 22, 2022 at 22:14

5$\begingroup$ We use Mathematica. We plan to make Vieta jumping implementation publicly available via Wolfram Community very soon. It solves many equations where standard FindInstance fails. Other parts of the program are currently not sufficiently clean and tested for the public release, but of course we plan to have everything public eventually. If you are interested to have untested version privately, and have access to Mathematica, you may email Aubrey. $\endgroup$– Bogdan GrechukCommented Apr 22, 2022 at 22:37

2$\begingroup$ Any chance to see an opensource implementation? $\endgroup$– Julien PuydtCommented Apr 23, 2022 at 7:36

2$\begingroup$ @Julien Puydt. We implement this is Mathematica. The Mathematica code will be public, but to run the code one needs Mathematica, which is not free. On the other hand, all algorithms are in the paper that is available for free from arXiv arxiv.org/abs/2108.08705 and most of the algorithms are not difficult, so anyone who is interested may implement them in any opensource computer algebra system they like. $\endgroup$– Bogdan GrechukCommented Apr 24, 2022 at 8:26

2$\begingroup$ one does not need nonfree Mathematica for nongraphics computations, Wolfram Engine is free to install and run for personal use. $\endgroup$– Dima PasechnikCommented Apr 24, 2022 at 8:46
2 Answers
There is no solution.
It is clear that at least one of $x$ and $y$ is positive and that neither is divisible by 7. We can assume that $a := x > 0$. The equation implies that there are integers $X$, $Y$ such that $$ X^2  7 a Y^2 = a (4 + a^3) $$ (with $X = a (a + 2y)$ and $Y = 2z$).
First consider the case that $a$ is odd. Then $4 + a^3$ is also odd (and positive), so we can consider the Jacobi symbol $$ \left(\frac{7a}{4+a^3}\right) \,. $$ One of the two numbers involved is ${} \equiv 1 \bmod 4$, so by quadratic reciprocity, $$ \left(\frac{7a}{4+a^3}\right) = \left(\frac{4+a^3}{7}\right) \left(\frac{4+a^3}{a}\right) = \left(\frac{4+a^3}{7}\right) $$ ($4 + a^3$ is a square mod $a$). Since $7 \nmid a$, we have $4 + a^3 \equiv 3$ or $5 \bmod 7$, both of which are nonsquares $\bmod 7$, so the symbol is $1$. This implies that there is an odd prime $p$ having odd exponent in $4 + a^3$ and such that $7a$ is a quadratic nonresidue $\bmod p$. This gives a contradiction (note that $p \nmid a$).
Now consider the case $a = 2b$ even; write $b = 2^{v_2(b)} b'$. Then we have that $4 + a^3 = 4 (1 + 2 b^3)$ and $$ \left(\frac{7a}{1 + 2b^3}\right) = \left(\frac{14b}{1 + 2b^3}\right) = \left(\frac{2}{1 + 2b^3}\right)^{1+v_2(b)} \left(\frac{7b'}{1 + 2b^3}\right) \,. $$ If $b$ is odd, then this is $$ \left(\frac{2}{1 + 2b^3}\right) (\left(\frac{1}{b}\right)) \left(\frac{1 + 2b^3}{7}\right) \left(\frac{1 + 2b^3}{b}\right) \,, $$ which is always $1$ (the product of the first two factors is $1$; then conclude similarly as above). We obtain again a contradiction.
Finally, if $b$ is even, then $$ \left(\frac{2}{1 + 2b^3}\right)^{1+v_2(b)} \left(\frac{7b'}{1 + 2b^3}\right) = \left(\frac{1 + 2b^3}{7}\right) \left(\frac{1 + 2b^3}{b'}\right) = 1$$ again (the first symbol is $1$, and quadratic reciprocity holds with the positive sign), and the result is the same.
Here is an alternative proof using the product formula for the quadratic Hilbert symbol.
If $(a,y,z)$ is a solution (with $a > 0$), then for all places $v$ of $\mathbb Q$, we must have $(7a, a(4+a^3))_v = 1$. We can rewrite the symbol as follows. $$ (7a, a(4+a^3))_v = (7, a (4 + a^3))_v (a, a)_v (a, 4+a^3)_v = (7, a(4 + a^3))_v $$ (the last two symbols in the middle expression are $+1$). So it follows that $$ (7, a)_v = (7, 4 + a^3)_v \,.$$
When $v = \infty$, the symbols are $+1$, since $a > 0$.
When $v = 2$, the symbols are $+1$, since $7$ is a $2$adic square.
When $v = p \ne 7$ is an odd prime, one of the symbols is $+1$ (and therefore both are), since $a$ and $4 + a^3$ have no common odd prime factors.
Finally, when $v = 7$, the symbol on the right is $$ (7, 4 + a^3)_7 = \left(\frac{4 + a^3}{7}\right) = 1 $$ as in the first proof.
Putting these together, we obtain a contradiction to the product formula for the Hilbert symbol.

2$\begingroup$ As is probably clear to Michael, this proof can be rewritten in terms of a BrauerManin obstruction to the integral Hasse principle coming from the quaternion algebra $(7a,a(4 + a^3))$ in the Brauer group of the surface. $\endgroup$ Commented Apr 24, 2022 at 7:43

1$\begingroup$ Thank you! What I like about the first proof is that it is completely elementary, and requires nothing beyond the definition and basic properties of the Jacobi symbol. $\endgroup$ Commented Apr 24, 2022 at 8:14

1$\begingroup$ @DanielLoughran Of course. Hilbert symbols and quaternion algebras are more or less the same thing. (The algebra used is actually $(7, 4+a^3) = (7, a)$.) $\endgroup$ Commented Apr 24, 2022 at 10:55

1$\begingroup$ So glad to see that computers will never replace humans 100% :) $\endgroup$– WolfgangCommented Apr 27, 2022 at 9:59

1$\begingroup$ @BogdanGrechuk In principle, you can make a list of all possible quaternion algebras (they are given by pairs of rational functions, which form a countable set) and check them one by one until you (or your computer) get(s) tired. This would not help much for concrete problems, of course. For certain classes of equations, one can describe the 2torsion of the Brauer group sufficiently explicitly to make this a more sensible approach. (One can also work with elements of higher order; for example, elements of order 3 can be used to deal with diagonal cubics in 4 variables.) $\endgroup$ Commented Apr 27, 2022 at 10:57
Here is a short proof, inspired by Michael Stoll's nice arguments.
Assume that $x,y,z\in\mathbb{Z}$ solve the equation. Then $xy(x+y)$ is positive and not divisible by $7$. Without loss of generality, $x$ is positive and not divisible by $7$. Using the identity $$\left(x^2 + \frac{2}{x+y}\right)^2 + 7\left(\frac{2z}{x+y}\right)^2 = x (4 + x^3),$$ we see that $x(4+x^3)$ is a norm in $\mathbb{Q}(\sqrt{7})$. Hence both $x$ and $4+x^3$ are norms in $\mathbb{Q}(\sqrt{7})$, because the only prime factor they can share is $2$, which splits in $\mathbb{Q}(\sqrt{7})$. In particular, $4+x^3$ is a quadratic residue mod $7$, contradiction.

4$\begingroup$ So glad to see that computers will never replace humans 100% :) $\endgroup$– WolfgangCommented Apr 27, 2022 at 9:59