Show that points on an elliptic curve have order 4

I am studying elliptic curves using this book and have a problem with task 4.11 which goes as follows:

Let $F_q$ be a finite field of odd characteristic and let $a,b \in F_q$ with $a \ne2b$ and $b \ne 0.$ Define the elliptic curve E by $y^2 = x^3+ax^2+b^2x$ a) Show that the points $(b,b\sqrt{a+2b})$ and $(-b,-b\sqrt{a-2b})$ have order 4.

So, if the curve was on normal weierstrass form I guess I could use the formula and eventually reach infinity, but if the curve can have characteristic 3, then how can I transform it to weierstrass? Or is there any other way to show that the points have order 4 using schoof's or baby-step giant-step etc?

• the characteristic of the field doesn't really play a role, just take the formula for adding points and compute $2P,3P$ and $4P$, where $P$ is your point. – Ferra Apr 30 '16 at 11:40
• @Ferra If I understand OP correctly, they want to use a form of the group law that applies just to curves in Weierstrass form (which, to be clear, does not always exist for an elliptic curve over $\Bbb F_3$). Also, one can just compute $2P$ and then $4P = 2P + 2P$. – Travis Willse Apr 30 '16 at 11:46
• to me, that form is called Weierstrass... probably he meant short Weierstrass form. In that case, when the characteristic is $3$ you can compute manually $2P$ and check that this is $2$-torsion. – Ferra Apr 30 '16 at 12:24

First, the form $y^2=x^3+ax^2+bx$ is close enough to the classical form to do all the necessary computations in, as long as the characteristic is not $2$.
Second, when an elliptic curve is in this form, I hope it’s obvious that the points of order two are exactly those on the $x$-axis.
Third, on this particular curve, the origin is thus a point of order two, and to find $2$-division of this point, you only need to find points on the curve at which the tangent passes through the origin.