How can I prove that $6y^2 = x(x + 1)(2x + 1)$ is a Weierstrass equation of an elliptic curve over rational numbers How can I prove that $6y^2 = x(x + 1)(2x + 1)$ is a Weierstrass equation of an elliptic curve over rational numbers ? How can I also find how many 2-torsion points has ?
 A: Substitute $x\to x-1/2$ to obtain $3y^2=x^3+ax+b$, substitute $y\to 3y, x\to 3x$ to obtain $y^2=x^3+Ax+B$. If $2P= O$ then $P= -P$ thus $y_P=0$, thus the 2-torsion corresponds to the roots of $x^3+Ax+B$ and the point at $\infty$.
A: The given equation is explicitly:
$$
6y^2=2x^3+3x^2+x\ .
$$
After a division by $2$ and multiplication with $3^3$, we get:
$$
(9y)^2 = (3x)^3+ \frac 92(3x)^2+\frac 92(3x)\ .
$$
On the R.H.S. isolate by force $X^3=\left(3x+\frac 32\right)^3$, so that the terms in $x^3, x^2$ are fully involved. It is then natural to substitute $Y=9y$, $y=\frac Y9$ and $X=3x+\frac 32$, $x=\frac X3-\frac 12$. The equation in $X,Y$ is then:
$$
Y^2=X^3-\frac 94X\ .
$$
The $2$-torsion points are the three points of order two, $(\pm 3/2,0)$, $(0,0)$, corresponding to the zeros of the R.H.S. above, together with the infinity point, $O$.
They are moved back in the $(x,y)$-world to the points 


*

*$(0,0)$ (obtained from $(3/2,0)$), 

*$(-1,0)$ (obtained from $(-3/2,0)$), 

*$(-1/2,0)$ (obtained from $(0,0)$). 


(Their first components correspond to the roots of $x(x+1)(2x+1)$.)
