Is equation $xy(x+y)=7z^2+1$ solvable in integers?

Question

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.

Answer 1

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.

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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.

Answer 2

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.