Q(\sqrt{-l_0}) satisfies Heegner hypothesis for an Elliptic curve of conductor C implies C is a square Trying to understand the proof of Corollary 2.3 in the following paper, 
http://arxiv.org/pdf/1312.3884.pdf
I would like to be able to justify that the root number of the quadratic twist $E^{(-l_0)}$ is $-w_E$, where $w_E$ is the root number of $E$. To spare the readers some time, the set up is the following
$E$ is an elliptic curve over $\mathbb Q$ with conductor $C$ and $K=\mathbb{Q}(\sqrt{-l_0})$, where $l_0>3$ is a prime congruent to $3$ modulo $4$. It is given that $K$ satisfies Heegner hypothesis for $E$, namely every prime factor of $C$ splits in $K$. Does this imply that $C$ is a square or something? If yes, then I can prove my assertion about the root number.
 A: It is a general fact that if $K$ is a number field, $K'/K$ is a quadratic extension, $E \rightarrow \mathrm{Spec}\, K$ is an elliptic curve, and $E' \rightarrow \mathrm{Spec}\, K'$ is its quadratic twist by $K'/K$, then the global root numbers are related by
$$w(E_{K'}) = w(E)w(E'),$$
even though the corresponding formula for local root numbers fails (there is an extra term $(-1, K'/K)$ that disappears globally due to the product formula). This may be derived from standard properties of root numbers or, in your situation, from looking at $L$-functions (I can provide you with a reference, if you want).
Once you have the above formula, it remains to note that $w(E_{K'}) = -1$ because there is a single infinite place and all the "bad" finite places come in pairs (due to the Heegner hypothesis), so the corresponding local root numbers kill each other in the product.
See also the related question Root number of a quadratic twist of an elliptic curve, whose answers also settle your present question.
