Converting equation into short Weierstrass form

Question

I have the curve $y^2 + 4y = x^3 + 3x^2 −x + 1$. I need to find a transformation of the form $X=x+a$ and $Y=y+b$ that turns this curve into the standard form of the elliptic curve: $$Y^2=X^3+AX+B.$$ I completed the square for the left side and now have $(y+2)^2= x^3 + 3x^2 −x + 5$, but I'm not sure what to do next to fix the right side. Any help would be great!

Answer 1

You need to complete the cube in terms of $x$, this is very similar to completing the square, except that we have $$(x+a)^3 = x^3 + 3ax^2 + 3a^2x + a^3$$ so if the coefficient of $x^2$ is $c$ then $(x + c/3)$ is the variable you should rewrite $x^3 + cx^2 +dx + e$ in terms of so that you get $$x^3 + cx^2 +dx + e =(x+c/3)^3 + A(x+c/3) + B$$ for some coefficients $A,B$ that you can solve for using this equation, in your case $c/3=1$ so this should work out nicely!

Answer 2

Hint: Observe that \begin{align} && y^2+4y&=(y+2)^2-4 \\ &\text{and}& \\ &&\qquad x^3+3x^2&=(x+1)^3-3x-1 \end{align} Can you continue?