<s>[INST] Converting equation into short Weierstrass form 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!
 [/INST] 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?
</s>