= Solution
For good reduction at $p$, the <reduction of an elliptic curve> gives a map
$$
E(\mathbb Q_p)\longrightarrow\widetilde E(\mathbb F_p)
$$
whose kernel is its <formal group of an elliptic curve>, and the torsion in that kernel is $p$-primary. Hence the prime-to-$p$ part of any torsion subgroup injects into $\widetilde E(\mathbb F_p)$.
At $p=2$, the odd part of $E(\mathbb Q)_{\rm tors}$ must divide $4$, so it is trivial. The torsion group is therefore a $2$-group. At $p=3$, all of this group has order prime to $3$ and injects into a group of order $7$, so it too is trivial. Thus
$$
E(\mathbb Q)_{\rm tors}=0.
$$
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