Is $y^2=x^3+7$ unsolvable modulo some $n$?

Question

The equation $y^2=4x^3+7$ has no integral solution since $y^2\equiv4x^3+7\pmod4$ has no solution (i.e. has no solution in $\Bbb{Z}/4\Bbb{Z}$).

It is well known that $y^2=x^3+7$ has no integral solution, but is there an integer $n (n>1)$ such that $y^2\equiv x^3+7\pmod n$ has no solution? If not, how can I prove it? Thanks in advance.

Answer 1

The discriminant of $x^3+7$ is $-27\cdot49$. Given a prime power $q$, this discriminant is non-zero in the field $F_q$ unless $q$ is $3,9,27,7,$ or $49$. For other values of $q$, $y^2=x^3+7$ is an elliptic curve over $F_q$, and we can use Hasse's bound

$$|N-(q+1)| \le 2\sqrt q$$

where $N$ is the number of points on the curve. So there are at least $q+1-2\sqrt q$ points on the curve. One of these is the point at infinity; subtracting this, we have at least $q-2\sqrt q$ solutions of $y^2=x^3+7$. This is $> 0$ (and therefore $\ge 1$) for all $q \ge 5$.

So you need only check the cases $q=2,3,4,7,9,27,$ and $49$ to prove that $y^2=x^3+7$ has a solution modulo every prime power. And as Greg Martin points out in a comment, this is enough to establish it for all $n$.

Answer 2

The congruence $y^2 \equiv x^3 + 7 \pmod n$ has a solution for every $n$. A proof involving Gauss sums is outlined in Exercise 6, pages 21 and 22, of the online document Characters.