I have equation of elliptic curve in Montgomery form over $F_p$.
E: $ y^2 = x^3 + Ax^2 + x$
I have to show that E has exactly three points of order 2 if discriminant $A^2 - 4$ is quadratic residue.
I've done some calculations and here is where I'm stuck: I've done implicit differentiations and got to the part where I have equation $x^3+Ax^2+x=0.$ Obviously one point is $(0,0)$ and two other points i can get by solving equation $x^2+Ax+1=0$ I can see for this to have two different solutions discriminant $A^2−4$ cannot be equal to $0$. But I can't see connection with that and the fact that $A^2−4$ is quadratic residue. I know definition of quadratic residue but not exactly what it brings me to this problem.
If $A^2-4$ is a (non-zero) quadratic remainder, then there exists an integer $m$, not divisible by $p$, such that $$ m^2\equiv A^2-4\pmod p. $$ We can view $m$ and $A$ as elements of $\Bbb{F}_p$, when the above congruence becomes an equality. This means that $m$ can handle the duties of $\sqrt{A^2-4}$ in the formula for the solutions of the quadratic equation $$ x^2+Ax+1=0, $$ that is $$ x=\frac{-A\pm\sqrt{A^2-4}}2=\frac{-A\pm m}2. $$ So if such an $m\neq0_{F_p}$ exists, you have found the three points on the curve with vertical tangents. If no such $m$ exists, then the quadratic formula tells us that $x^2+Ax+1$ has no zeros in the field $\Bbb{F}_p$. When that happens, the origin is the only point on your curve with a vertical tangent. Meaning that there is a single 2-torsion point.