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I do not see why your statement is true in the following example. Take $\psi$ to be the $2$-isogeny from $E : y^2=x^3+x$ to the non-isomorphic $E': y^2 = x^3+x+3$ killing the point $(2,0)$. Now the image under the dual $\hat\psi$ of $I(E,E')$ in $End(E)$ is the ideal generated by $[2]$. I believe that one concludes from this that any isogeny $\varphi:E\to E'$ has to factor through $\psi$. So $H=<(2,0)>$ vanishes for all $\varphi$
In other words: There is a lot of research available on integral points on elliptic curves and the resulting algorithm is implemented in sage and magma. See for instance chapter XIII of Smart's "Tha Algorithmic Resolution of Diophantine Equations". The sage documentation of this function refers to Petho A., Zimmer H.G., Gebel J. and Herrmann E., Computing all S-integral points on elliptic curves Math. Proc. Camb. Phil. Soc. (1999), 127, 383-402.
If the analytic rank of an elliptic curve over $\mathbf{Q}$ is 1 then the height of the Heegner point is a period by Gross-Zagier, I believe. But I don't think anything like this is known for higher rank. But that is completely different from your question...
389: maybe I remembered the wrong thing. Christophe Delaunay computed the critical points for this modular parametrisation at the end of archive.numdam.org/ARCHIVE/JTNB/JTNB_2005__17_1/…. I will ask him, if he knows more. [PS: Yes, see you on Wednesday, looking forward to your talk.]
You won't need BSD: If the analytic rank of the twist of $E$ corresponding to $K$ were zero, then, by Kolyvagin, the rank of this twist would be zero, too.
If the mod-2 representation of the Jacobian $E$ of $C$ is surjective, then there are no $2$-torsion points on $E(K)$ for quadratic $K$. So the rank of $E$ has to grow in every quadratic $K/\mathbb{Q}$ and so must the analytic rank. But that is not possible by non-vanishing results, I believe. Do I miss something ?
I think there is a misunderstanding here. The coefficients of the formal group law of an elliptic curve are not the same as the coefficients of the L-series. Your $f$ is $g(q)/q$ for the associated newform $g$. I don't see why this should give a formal group.
@jerr18 this is ONE minimal model, there are others. sage, magma etc transform futher such that $a_1$, $a_2$, $a_3$ are 0,1, or -1. It is easy to do this depending on the parity of $n$. Tell me if you get a different model if you start from my equation.
