I'm confused by what appears to be contradictory information.
In this post, the claim is made that
"Every elliptic curve over $\mathbb{Q}$ can be written in the form $y^{2}= x^{3}+ax+b$ where $a,b∈ \mathbb{Z}$ with discriminant $Δ=−16(4a^{3}+27b^{2})≠0$. So the number of elliptic curves of discriminant $D$ is bounded above by number of nontrivial pairs $(a,b)∈ \mathbb{Z}^{2}$ such that $D=−16(4a^{3}+27b^{2})$."
Is this true? this post gives an example of an elliptic curve with integer coefficients that is not isomorphic to any curve $y^2 = x^3 + Ax + B$ for $A, B$ integers and discriminant equal to the original. Isn't this in contradiction to the original claim? That is, the curve in the second link would not be counted as a curve with its discriminant, right?
In fact, it seems to me (after plugging through the relevant changes of variables) that most curves of the form $$E: y^2 + a_{1} xy + a_{3} y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6},$$ with discriminant $\Delta$
cannot be written in the form $$ E' : y^{2} = x^{3} + Ax + B $$ with discriminant also $\Delta$ for some $A, B \in \mathbb{Z}$.
Any help?
The statement:
is certainly false as stated. For example, the discriminant of the model for the curve $$E:y^2 + y = x^3 - x^2 - 10x - 20$$ is $-11^5$. Any model of the form $y^2=x^3+Ax+B$ with integers $A,B$ has a discriminant of the form $D=-16(4a^3+27b^2)$. But the equation $$-11^5 = -16(4a^3+27b^2)$$ is impossible in integers $a,b$ as the left hand side is odd and the right hand side is even.
One could salvage the statement above, however, and write something in a similar spirit. Going from a generic Weierstrass model to a short Weierstrass model means that the discriminant is multiplied, at worst, by $2^{12}3^{12}$. So one could say
and I believe this is a true statement.