Let $E$ be an elliptic curve and $E\_n$ be its quadratic twist by $n$. Let $\phi\_n: X\_0(N\_n) \to E\_n$ be the normalized modular parametrization of $E\_n$ ($\infty \to O$)


For some particular curves $E$, it seems (based on computing some examples in sage) to always be the case that $\phi\_n(0) = O$ when the rank of $E\_n$ is 0. For the curves I'm looking at, the torsion subgroup is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ for all of the $E\_n$. What are some possible explanations for this or strategies to go about proving it? 


Note: Edited to clarify the first comment. 



Possible reasons why the image of 0 in the modular parametrization would always be O for a family of quadratic twists of elliptic curves
Note: this gives a plausible explanation, not a whole solution.


Hi, the BSD-quotient is $L(E\_n,1)/\Omega\_n = Sha\cdot\prod\_p c\_p(E\_n)/|T|^2$. I think your question is mostly equivalent to asking why this quotient is an integer when twisting by $n$? The torsion is of size 4 (maybe there is an exceptional twist where it is larger, you did not specify). I think because of the full 2-torsion, the Tamagawa number should be 4 at any prime dividing $n$, except those dividing the original conductor. So if $n$ has at least two prime factors coprime to $N$ it seems you are done. When twisting by a prime or some $n$ with nontrivial gcd with $N$, then you need to take a more careful analysis, but the original curve itself should already have nontrivial Tamagawa numbers at some prime if indeed it has full 2-torsion. In short, the product of Tamagawa numbers outweighs the torsion in the denominator.


The definition of "0" here is just the cusp 0 on the boundary of the complex upper half plane. One is essentially integrating the modular form as $\int\_0^{i\infty} f(z) dz$, to get the modular parametrization. But there are the constants like $2\pi$ and the Manin constant to worry about.


