Elkies proved The existence of infinitely many supersingular primes for every elliptic curve over Q. I read his paper, but found the supersingular primes he constructed are all 3(mod 4) type. So, how about the others?
2 Answers
It's the auxiliary prime $l$ that must be $3 \bmod 4$; the supersingular prime $p$ is not guaranteed to be congruent to $3 \bmod 4$, and indeed the residue of $p \bmod 4$ is unpredictable (unless the curve has CM by an order in ${\bf Q}(i)$). Several examples on page 565 have $p \equiv 1 \bmod 4$. For example, the curve $X_0(11)$: $y^2+y=x^3x^2$ has $j = 2^{12}/11$ and bad reduction only at $11$, so we can take $l=7$ because the Legendre symbol $(11/7)$ is $+1$. The polynomial $P_l(X)$ is then $P_7(X) = X + 15^3$, and we find $P_7(j) = 33029/11$, with $(33029/7) = 1$. Hence some prime factor of $33029$ is supersingular. It so happens that $33029$ is itself prime, and congruent to $1 \bmod 4$.

$\begingroup$ I see. I should read the paper more carefully. I now have another question, mathoverflow.net/questions/149781/… . Maybe you have considered. $\endgroup$ Nov 24, 2013 at 6:16

$\begingroup$ But I read an article mentioning the bias appearing among supersingular primes in CONGRUENCE CLASSES when consider in an average sense thesis.library.caltech.edu/6242/1/Nahid_Waljithesis.pdf . May it be so in the LT Conecture? Can We apply the same method as in LT's article to get a constant for a CONGRUENCE CLASSES OF PRIMES $\endgroup$ Dec 9, 2013 at 5:52
Having ordinary or supersingular reduction at $p$ for CM elliptic curves depends on the splitting type of $p$ in $\mathrm{End}(E) \otimes \mathbf{Q}$, see reduction of CM elliptic curves

$\begingroup$ So a curve like $y^2=x^3x$, for instance, will only have supersingular reduction at $p\equiv 3$ mod $4$. $\endgroup$ Nov 23, 2013 at 17:21

4$\begingroup$ @Timo: For CM curves, one knows what's going on. Elkies paper is really about nonCM curves. I hadn't realized that the infinitely many ss primes he constructs are all 3 (mod 4). For a nonCM curve over $\mathbb{Q}$, I see no reason why there shouldn't be infinitely many ss primes in any arithmetic progression (containing infinitely many primes, of course). But possibly there are some local constraints. Maybe Noam can chime in. $\endgroup$ Nov 23, 2013 at 17:26