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?
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. 


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 semisimplification of $\rho$ is $1 \oplus 1$ (a semisimple representation of dim 2 is characterized by its trace and determinant, even in characteristic 2). Therefore $\rho$ fixes a line, and the nonzero point of this line is a rational 2torsion point of $E$. 

