Solutions to $y^2 = x^3 + x$ in $\mathbb{P}^2(\mathbb{F_p})$.

Question

Consider the elliptic curve $$\mathcal{C}: y^2 = x^3 + x.$$

Let us consider the reduction of $\mathcal{C} \mod{p}$.

Some explicit computation show that $\# \tilde{\mathcal{C}}(\mathbb{F}_3) = 4$, $\# \tilde{\mathcal{C}}(\mathbb{F}_5) = 4$, $\# \tilde{\mathcal{C}}(\mathbb{F}_7) = 8$, and $\# \tilde{\mathcal{C}}(\mathbb{F}_{11}) = 12$ (these totals include the point at infinity).

On page 138, Example 4.6, Silverman and Tate go on to say that it is "not hard to check that" $4 | \# \tilde{\mathcal{C}}(\mathbb{F}_p)$ for all $p \geq 3$.

However, I am struggling to verify this fact myself, and would be very grateful for any assistance!

From working the smaller cases by hand, it seems like a natural divide would be between $p \equiv 1 \mod{4}$ and $p \equiv 3 \mod{4}$. However, if feels like there must be some theorems about solution in these two separate cases that I am missing right now.

Answer 1

Yes, a simple way is to split the argument into two cases according to residue class of $p$ modulo $4$.

• If $p\equiv3\pmod4$ then $-1$ is not a quadratic residue. So for all $x\neq0$ either $x$ or $-x$ gives rise to two points while the other gives no points. It follows that the number of points on this curve is exactly $p+1$, and hence divisible by four. See this question for the details.
• If $p\equiv1\pmod4$ then $-1$ is a quadratic residue. Implying that $x^3+x=x(x^2+1)$ has three distinct zeros in $\Bbb{F}_p$. But the zeros of the cubic are all points of order two on the curve, so the $2$-torsion subgroup $E[2]\le E(\Bbb{F}_p)$ has at least four elements (those three and the point at infinity). Actually it cannot have more than four elements, but for the task at hand it is enough that there are at least four, all in $E(\Bbb{F}_p)$. So $4\mid \#E[2]\mid \# E(\Bbb{F}_p)$ and we are done.