Heegner Points on $X_0(N)$ when some primes dividing $N$ are inert in the imaginary quadratic field If $K = \mathbb{Q}(\sqrt{-D})$ is a imaginary quadratic field with discriminant $-D$, then we get Heegner points on $X_0(N)$ as long as there exists $\mathfrak{n} \subset \mathcal{O}_K$ such that $\mathcal{O}_K/\mathfrak{n} \simeq \mathbb{Z}/N$.  For this to happen, we need all of the primes $p \mid N$ to either split or ramify in $\mathcal{O}_K$.  Gross-Zagier was originally proved when the "Heegner hypothesis" is satisfied, i.e. all of the primes $p$ split.  However, in applications, this hypothesis is not always satisfied.  
However, Heegner's original paper used $D \equiv 3 \pmod{8}$ and $N = 32$, so $2 \mid N$ is inert.  How do you get points on $X_0(N)$ in this situation? I couldn't find an English translation or summary of Heegner's paper, and am confused as to how this works. 
 A: I think it historically known that Heegner only used "mock" Heegner points, see papers of Monsky (see page 46:
"It should be noted that the modular points that Heegner uses to show that $2p_3$ is a congruent number aren't "Heegner points" either; they arise from the field $Q(\sqrt{ip_3})$ in which 2 is inert."
In the paragraph below that, Monsky gives an outline of the method (he is unifying various cases, including Heegner's). He works on $2Y^2=X^4+1$ rather than $E$ directly. He does not use the language of modular curves per se, but I suspect it can be made equivalent. Perhaps if one does so, there could also be "partial traces" from a covering modular curve, corresponding to genus theory when summing over quadratic forms.
http://link.springer.com/article/10.1007%2FBF02570859
Edit: here is what I mean in the later paragraph. The idea dates to Birch and Stephens, and Darmon, Tornaria and Pacetti use it more recently (for Waldspurger, not always Heegner). See 2.1.3 (Twisted traces) of [1]. You can leverage information about inert primes, by twisting. I do not handle all the details, but I perceive this is how Heegner works.
The idea is that the Hilbert class field of $Q(\sqrt{-dl})$ contains that of $Q(\sqrt{-d})$, and if we construct a point in the former, then tracing by the genus character $\chi_l$ gives a point in $Q(\sqrt{-d})$. An explicit example is then given, where $-d=-3$ for 76A, and this is not square. You twist by $l=5$, to leverage.
[1] http://jtnb.cedram.org/item?id=JTNB_2008__20_3_829_0
