Number of points of elliptic curve $y^2 = x^3 - x$ over $\mathbb{F}_{p^2}$ with $p \equiv 3 \mod 4$

Question

Let $p$ be a prime number with $p \equiv 3 \mod 4$, and consider the elliptic curve $$ E/\mathbb{F}_{p^2}: \quad y^2 = x^3 - x. $$ I suspect that $E(\mathbb{F}_{p^2})$ has order $(p+1)^2$ as I observed it with MAGMA (the code I used can be found below).

I found a similar-looking argument in Jyrki Lahtonen's answer here (where he dealt with the same curve over $\mathbb{F}_p$ instead of $\mathbb{F}_{p^2}$). It seems that the crucial argument there was that $-1$ is not a quadratic residue modulo $p \equiv 3 \mod 4$. However, I think $-1$ will become a square in $\mathbb{F}_{p^2}$, so the argument fails I think.

Question How can I show my claim instead?

Appendix: My MAGMA code (Output gives only "true").

L := [ x : x in [3 .. 100] | IsPrime(x) and x mod 4 eq 3];

for p in L do
    K := GF(p^2);
    R<x> := PolynomialRing(K);
    f := x^3 - x;
    E := EllipticCurve(f);
    #E eq (p+1)^2;
end for;

Answer 1

For $\mathbb{F}_p$, the argument is that $$\#E(\mathbb{F}_p) = 1 + \sum_{x\in \mathbb{F}_p} \left(1+{{\genfrac{(}{)}{}{}{x^3-x}{p}}}\right)$$

Since $x^3-x$ is an odd function, and $p \equiv 3 \pmod{4}$, it follows that ${{\genfrac{(}{)}{}{}{x^3-x}{p}}}$ is also an odd function, so $$\sum_{x\in \mathbb{F}_p} {{\genfrac{(}{)}{}{}{x^3-x}{p}}} = 0$$ which implies $\#E(\mathbb{F}_p) = p+1.$

For $\mathbb{F}_{p^2}$, the easiest way to see that $\#E(\mathbb{F}_{p^2}) = (p+1)^2$ is to take the complex multiplication endomorphism $\phi\colon E \to E$ sending $(x,y)$ to $(-x,y\cdot \sqrt{-1})$. From standard theory, $$\#E(\mathbb{F}_p) = \deg(1-\pi)$$

$$\#E(\mathbb{F}_{p^2}) = \deg(1-\pi^2)$$ where $\pi$ denotes the $p$-th power Frobenius map. But $1-\pi^2 = (1-\pi)(1+\pi)$, and $\deg(1+\pi) = \#E(\mathbb{F}_p)$ because $1+\pi$ is separable and the kernel of $1+\pi$ is equal to the image of $E(\mathbb{F}_p)$ under $\phi$, so $$\deg(1-\pi^2) = \deg(1-\pi)\deg(1+\pi) = (p+1)^2.$$

For a general elliptic curve $E: y^2 = x^3 + ax + b$ and $p \equiv 3 \pmod{4}$, a version of this argument still works. The map $\phi$ as defined above maps from $E$ to $E' : y^2 = x^3 + ax - b$ and we have $\#E'(\mathbb{F}_p) = p+1+t$ where $t$ is the trace of $E/\mathbb{F}_p$, so we obtain $$\#E(\mathbb{F}_{p^2}) = (p+1-t)(p+1+t).$$ Alternatively, the proof of rationality of the zeta function for $E$ immediately gives you the above formula in full generality, as well as formulas for $\#E(\mathbb{F}_{p^k})$ for all $k$.