= Solution
Modulo five the curve is $y^2=x^3+x$, with <good reduction> because $4a^3+27b^2\equiv4\pmod5$. It has one point above each of $x=0,2,3$ and no others, plus infinity, so its group has order four. Rational torsion therefore has order dividing four.
A nonzero <rational point> of order two would have $y=0$ and integral $x$, so would give an <integer> root of $x^3+ax+b$. But modulo two this <polynomial> is $x^3+x+1$, which is one at both residues. Thus there is no rational <2-torsion>. Every nontrivial finite group of order dividing four contains an element of order two, so
$$
\boxed{E(\mathbb Q)_{\rm tors}=\{O\}.}
$$
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