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?



Elliptic curves with trace of Frobenius values always congruent to 0 modulo 2
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**.


