Now that I'm thinking about it, if $\mathfrak{p}\cap \mathbf{Z}=p\mathbf{Z}$ is inert in $K$ then, for $p\geq 5$, the eigenvalues of $Fr_{\mathfrak{p}}$ are associated algebraic numbers which must differ by a sign, and therefore the coefficien $Tr(Fr_{\mathfrak{p}})=a_{\mathfrak{p}}=0$. Therefore, $\tilde{E(O_L/\wp)$ has size p^2+1 which is coprime to $p$.

If E is defined over $Q$ and has CM by $K$, then E will be supersingular at p if and only if p is inert in K. Does it help ?

Have a look at p. 116 of the document above. I only rewrote Ehrenfeucht's proof with a few more details. I can send you the original paper tomorrow, if you like.

Dear Georges, have a look at the following mat.ulaval.ca/fileadmin/mat/documents/Chapdelaine/Algebre_II‌​I/…

I have in mind the book of Lax and Philips who applied scattering theory to the study of automorphic forms. In that book I saw for the first time a truncated version of the hyperbolic Laplacian which has a compact resolvent. Then from a key observation of Colin de Verdiere, one may apply the Fredholm analyticity theorem to obtain the analytic continuation and the functional equation. In any case, thanks again for your answer.