<s>[INST] Rewriting elliptic curve with point of order 3 I have a question regarding a change of coordinates with elliptic curves. I tried to show that an elliptic curve $E: y^2 = x^3 + ax^2 + bx +c$ with coefficients in $\mathbb{F}_q[t]$ and where $P$ is a rational point of order 3 can be rewritten as:
\begin{equation*}\label{}
E\colon y^2 = x^3 + A(x-B)^2,
\end{equation*}
where the coefficients $A$ and $B$ are from $\mathbb{F}_q[t]$. So I let $P=(\alpha,\beta)$ be a $\mathbb{F}_q(t)$-rational point of order $3$ on an elliptic curve given by $E: y^2 = x^3 + ax^2 + bx + c$. Shifting the $x$-coordinate of $P$ to $0$, i.e. replacing $x$ by $x+\alpha$ gives:
\begin{equation*}\label{}
\begin{split}
y^2&= (x+\alpha)^3 + a(x+\alpha)^2 + b(x+\alpha) + c \\
&= x^3 + x^2(3\alpha + a) +x(3\alpha^2 + 2a\alpha+b)+\alpha^3 + a\alpha^2 + \alpha b +c \\
&= x^3 + x^2(3\alpha + a) +x(3\alpha^2 + 2a\alpha+b)+\beta^2.
\end{split}
\end{equation*}
Renaming $3\alpha + a$ to $a$ and $3\alpha^2 + 2a\alpha+b$ to $b$ gives an equation of the form: 
\begin{equation*}\label{}
y^2 = x^3 + ax^2 + bx + \beta^2.
\end{equation*}
The point $(0,\beta)$ is of order $3$, hence we need $2P = -P$. Using the duplication formulas we obtain that $\left(\frac{b^2-4a\beta^2}{4\beta^2},\frac{4ab\beta^2 - b^3 - 8\beta^4}{-8\beta^3}\right) = (0,-\beta)$ and hence $b^2 = 4a\beta^2$. Suppose that $b \neq 0$, then $a\neq 0$ and $a$ must be a perfect square. All of this means that we can take an element $z \in \mathbb{F}_q(t)$ such that $z^2 = \frac{\beta^2}{a}$. Combining everything yields the following equation:
\begin{equation*}\label{}
\begin{split}
y^2 &= x^3 + ax^2 + \sqrt{4a^2z^2}x + z^2a \\
&= x^3 + ax^2 + 2azx + z^2a \\
&= x^3 + a(x+z)^2.
\end{split}
\end{equation*}
Here is where I don't know how to continue. I can't seem to find a way to force $z$ to be an element of $\mathbb{F}_q[t]$. Any help?
 [/INST] Let us write $z=f/g$ with polynomials $f,g\in\Bbb F_p[t]$. We multiply with $g^6$ the last equation $y^2=x^3+a(x+z)^2$ getting:
$$
(yg^3)^2 = (xg^2)^3 + ag^2(xg^2+fg)^2\ ,
$$
and use the new "coordinates" $X = xg^2$, $Y=yg^3$.
Note: $y^2=x^3+C^2$ is not of the wanted form. It has the torsion points $T_\pm=(0,\pm C)$ of order $3$.
Note: If the constant $a$ in $y^2=x^3+a(x+z)^2$ has a denominator $d$, we multiply with $d^6$, group $y^2d^6=(yd^3)^2$, and $x^3d^6=(xd^2)^3$, then push from $ad^6=ad^2\cdot d^4$ the factor $d^4$ inside the square, so that $(xd^2)$ also appears inside the square.
</s>