TITLE: Is equation $xy(x+y)=7z^2+1$ solvable in integers?
QUESTION [31 upvotes]: 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.

REPLY [20 votes]: 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.