= Solution
For a nonsingular curve $y^2=x^3+Ax+B$ with <integers> $A,B$, the <Nagell–Lutz theorem> states that a finite <torsion subgroup>[rational torsion] point has <integer> coordinates and satisfies either $y=0$ or
$$
\boxed{y^2\mid4A^3+27B^2.}
$$
The point at infinity is the identity and is treated separately. Integrality is the result allowed in the source. To prove the remaining divisibility, suppose $y\ne0$. Then the doubled point is finite and also torsion, so its coordinates are integral. Its <tangent> slope is $m=(3x^2+A)/(2y)$ and its abscissa is $m^2-2x$. Thus $m^2$ is an <integer>. A <rational number> whose square is an <integer> is itself an <integer>, so $2y$ divides $3x^2+A$. In particular $y^2$ divides $\psi_1(x)=(3x^2+A)^2$. It also divides $\psi_2(x)=x^3+Ax+B=y^2$. Evaluating the permitted <polynomial> identity at the <integer> $x$ therefore proves the divisibility of $4A^3+27B^2$ by $y^2$.
For $y=0$, the point equals its negative and has order two. The theorem gives a finite set of torsion candidates, but the divisibility condition by itself does not assert that every candidate is torsion.
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