Elliptic curves with trace of Frobenius values always congruent to 0 modulo 2 Let $E/\mathbb{Q}$ be an elliptic curve and suppose that the trace of Frobenius values are such that $a_{p}(E) \equiv 0 \pmod{2}$ for all odd primes avoiding the conductor. Is it the case that $E$ contains a nontrivial rational two torsion point? Why?
 A: Alternative method without using the cubic equation (of course not fundamentally different): the Galois representation $\rho: G_{\mathbb Q} \rightarrow GL_2(\mathbb Z/2 \mathbb Z)$ on the points of $2$-torsion of $E$ satisfies tr $\rho(Frob_p)=0$ for every odd prime $p$ by hypothesis, hence tr $\rho=0$ by Chebotarev. (edited:) It has obviously determinant $1$. Hence the semi-simplification of $\rho$
is $1 \oplus 1$ (a semi-simple representation of dim 2 is characterized
by its trace and determinant, even in characteristic 2). Therefore $\rho$ fixes a line, and the non-zero point of this line is a rational 2-torsion point of $E$.
A: Yes.  Let $P(x) \in {\mathbb Q}[X]$ be the cubic whose roots are
the $x$-coordinates of the $2$-torsion points.  The hypothesis says that
$P$ has a root mod $p$ for all but finitely many primes $p$.
If $P$ were irreducible then its Galois group would contain a $3$-cycle, 
and then there would be infinitely many $p$ such that $P$ remains 
irreducible mod $p$ (using Čebotarev).  Hence $P$ is reducible.
Since its degree is only $3$, it follows that $P$ has a rational root,
whence $E$ has a rational $2$-torsion point, QED.
