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Is there an "easy" way to count points on a twists of elliptic curves? Say we consider the curve $\newcommand{\F}{\mathbb F} E(\F_p):\ y^2=x^3+x$ and her twist $E'(\F_{p^2})=x^3+2^{1/4}x$. (Some may see, this is a curve from the KSS-16 family.) Our basic-setting is: We know p, $\#E$ and the trace of frobenius $t$.

Let me understand first some different things. If I take the definition of a degree of a twist (attached), as I know it, this is a quartic/degree four twist? The fourth root of 2 has degree 4.

Def: Twist of Degree {2,4,6}

Consider $y^2=x^3+ax+b$ then the twists are given by $y^2 = x^3+a_ix+b_i$

If $ab\neq 0$, then $a_i=a, b_i=b/c^i$ where $deg(c)=2$ and $i=1,2$

If $b=0$, then $a_i=a/c^i$, where $deg(c)=4$ and $i=1,...,4$

If $a=0$, then $b_i=b/c^i$, where $deg(c)=6$ and i=1,...,6$

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Shalec
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  • If $a$ is a quadratic residue modulo $p$ then $ay^2 = x^3+x$ is isomorphic to $y^2=x^3+x$. If $a$ is a quadratic non-residue then for each $x, x^3 +x \ne 0$ there is a $y$ such that $(x,y)$ is a point of $y^2 = x^3+x$ or $ay^2 = x^3+x$ but not both. For twists on the $x+ax+b$ side, you should be able to apply a similar argument. – reuns Aug 18 '17 at 15:24

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If $E$ is an elliptic curve over $\Bbb F_q$ with $p+1-t$ points, its quadratic twist has $p+1+t$ points.

Quartic twists only occur in curves with CM over $\Bbb Z[i]$. This multiplies the eigenvalues of Frobenius by powers of $i$. So if the trace of Frobenius is $t$, then $t=2a$ where $a\pm ib$ are eigenvalues of Frobenius. One of twisted curves has eigenvalues $b\pm ia$ and so has $p+1-2b$ points. The other has $p+1+2b$ points.

Cubic and sextic twists only occur over curves with CM over $\Bbb Z[\omega]$, $\omega=\frac12(-1+i\sqrt3)$. Similar considerations apply as in the quartic case, but multiplying by sixth roots of unity this time. I'll omit the details...

Angina Seng
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  • So this twist is a quadratic one? We could take $\mathbb F_{p^2}\cong \mathbb F_p(i)$. As you see, I don't get the connection at all. :) I see, that you gave me the orders of the possible twists and a way to compute them. – Shalec Aug 18 '17 at 13:24
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    If you know the traces of Frobenius for $y^2=x^3+x$ over $\Bbb F_p$ you can work them out over $\Bbb F_{p^2}$ by squaring them. Then you can work out the what they are for a curve $y^2=x^3+sx$ over $\Bbb F_{p^2}$, essentially by evaluating the quartic character of $s$. @Shalec – Angina Seng Aug 18 '17 at 14:56
  • There comes another question up: How did they get to this twist? Actually I thought, they were starting through $\mathbb F_p$ and find their twist in $\mathbb F_{p^2}$. But as it comes out, they consider $\mathbb F_{p^{16}}$ and goes down to $\mathbb F_{p^2}$, which makes this twist a octic (?) twist.

    If I consider $\newcommand\F{\mathbb F} \F_{p^{16}}\cong \F_{p^2}[x]/(x^8-\sqrt{2})$, where $\F_{p^2}\cong \F_p[y]/(y^2-2)$ then the base for twisting should be a solution of $x^8-\sqrt{2}=0$ ($\sqrt$ is literally used)

     – Shalec Aug 19 '17 at 14:36