 Solution

For a nonsingular curve $y^2=x^3+Ax+B$ with integers $A,B$, the Nagell–Lutz theorem states that a finite rational torsion point has integer coordinates and satisfies either $y=0$ orThe point at infinity is the identity and is treated separately. Integrality is the result allowed in the source. To prove the remaining divisibility, suppose $y\ne 0$. 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.