= Solution
Here $4A^3+27B^2=31$. A nonzero ordinate of a <torsion point of an elliptic curve>[torsion point] must have its square dividing 31, so $y=\pm1$. The curve equation then becomes $x(x^2+1)=0$, whose only <integer> solution is $x=0$. No point with $y=0$ exists over the rationals: the monic cubic $x^3+x+1$ has no <integer> <root of a polynomial>[root], since its only possible <integer> <roots of a polynomial>[roots] are $\pm1$ and neither works.
It remains to test $(0,\pm1)$. Doubling either gives abscissa $(1/(\pm2))^2=1/4$, which is not an <integer>. If the original point were torsion its double would satisfy the integrality part of the <Nagell–Lutz theorem>, a contradiction. Therefore
$$
\boxed{E(\mathbb Q)_{\mathrm{tors}}=\{\mathbf o\}.}
$$
The <rational point> $(0,1)$ consequently has infinite order, so its multiples give infinitely many distinct <rational points>.
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