Order of elliptic curve $y^2 = x^3 + ax^2 + b^2x$ is multiple of $4$.

Question

Let $\mathbb{F}_q$ be a finite field with odd characteristic and let $a,b \in \mathbb{F}_q$ with $a \neq \pm 2b$ and $b \neq 0$. Define the elliptic curve $E: y^2 = x^3 + ax^2 +bx$

The goal is to show that $4 \ | \ E(\mathbb{F}_q)$.

The first step is showing that the points $(b,b\sqrt{a+2b})$ and $(-b,-b\sqrt{a-2b})$ have order 4, which I was able to show.

Now, I want show the following things:

1. At least one of $a +2b,\ a-2b, \ a^2-4b^2$ is a square in $\mathbb{F}_q$ .
2. If $a^2-4b^2$ is a square in $\mathbb{F}_q$, then $E[2] \subseteq E(\mathbb{F}_q)$.

If I show these 2, it follows that $4 \ | \ E(\mathbb{F}_q)$ , however I fail to show these 2 properties.
Can I somehow get a contradiction by assuming that none of these is a square, or how is this done?

Answer 1

For the second part, consider this:

The roots of $x^2 + ax +b^2$ are given by $ t_{1,2} = \frac{-a \pm \sqrt{a^2-4b^2}}{2}$.

If $a^2-4b^2$ is a square in $\mathbb{F}_q$ then the above expression is in $\mathbb{F}_q$. Thus, $y^2 = x(x-t_1)(x-t_2)$ has 3 roots .