 Solution

Modulo five the curve is $y^2=x^3+x$, with good reduction because $4a^3+27b^2\equiv 4(\mathrm{mod}5)$. 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