= Solution
The <Lutz–Nagell theorem> says that if
$$
E:y^2=x^3+ax+b,
\qquad a,b\in\mathbb Z,
$$
has nonzero discriminant and $P=(x,y)\in E(\mathbb Q)$ is a torsion point, then $x,y\in\mathbb Z$ and either $y=0$ or
$$
y^2\mid4a^3+27b^2.
$$
For integrality, fix a prime $p$. If a rational point has nonintegral coordinates, its primitive projective coordinates reduce to $O$, so it belongs to the kernel of reduction. The parameter $t=-x/y$ identifies this kernel with the <formal group of an elliptic curve> over $p\mathbb Z_p$. The formal logarithm, with the standard separate first-step argument at $p=2$, shows that this group has no nonzero rational torsion. A rational torsion point therefore has nonnegative $p$-adic valuations in both coordinates for every prime $p$, hence integral coordinates.
Suppose now that $y\ne0$. The point $2P$ is again a nonzero torsion point and hence integral. Its $x$-coordinate is $m^2-2x$, where
$$
m=\frac{3x^2+a}{2y}.
$$
Thus $m^2$ is an integer. A rational number whose square is integral is integral, so $2y\mid3x^2+a$ and in particular $y^2\mid(3x^2+a)^2$. The curve equation and the identity
$$
(3x^2+4a)(3x^2+a)^2-27(x^3+ax-b)(x^3+ax+b)=4a^3+27b^2
$$
then prove $y^2\mid4a^3+27b^2$. This is the <divisibility proof in the Nagell–Lutz theorem>.
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