For most prime powers $q$, an elliptic curve over $\mathbb{F}_q$ has a point of order greater than $4\sqrt{q}$ This is Exercise $4.6f$ in Washington's book on elliptic curves.
Show that for most values of $q$, an elliptic curve over $\mathbb{F}_q$ has a point of order greater than $4\sqrt{q}$.
In this exercise, we write $E(\mathbb{F}_q) \cong \mathbb{Z}_n \oplus \mathbb{Z}_{mn}$ and in part $e$, it was shown that when $m \geq 17$ and $q$ is sufficiently large then $E(\mathbb{F}_q)$ has a point of order greater than $4\sqrt{q}$. However, I am not sure how to show this in general. I have tried to show that $m$ is almost always at least $17$ although I am not so sure that is actually true. In part $d$, it was shown that $mn \geq \sqrt{m}(\sqrt{q}-1)$ and this was used for the case where $m \geq 17$ although this doesn't seem to lead anywhere otherwise.
Any help is appreciated!
 A: Something is missing from the list of assumptions. Possibly only a precise description of "for most prime powers $q$". Below I describe a curve such that the claim is false for infinitely many choices of $q$.
Consider the case of $q=2^t$, $t$ even, say $t=2\ell$, and the elliptic curve $E$
$$
y^2+y=x^3.
$$
From this answer we see that the point doubling formula of this curve reads
$$
[2](\xi,\eta)=(\xi^4,\eta^4+1).\qquad(*)
$$
Iterating $(*)$ a bit and recalling the characterstic two quirk of $(a+b)^4=a^4+b^4$ we see that for every point $(\xi,\eta)\in E(\Bbb{F}_q)$ we have
$$
[2^\ell](\xi,\eta)=(\xi^{2^{2\ell}},\eta^{2^{2\ell}}+1)=(\xi,\eta+1)=[-1](\xi,\eta).
$$
This immediately implies that the order of an arbitrary point $(\xi,\eta)\in E(\Bbb{F}_q)$ is a factor of $2^\ell+1=\sqrt q+1$.

It is not difficult to show that here actually the group $E(\Bbb{F}_q)\simeq \Bbb{Z}_{2^\ell+1}\oplus\Bbb{Z}_{2^\ell+1}$. In other words, $n=2^\ell+1$ and $m=1$. After all, this is the only two generator finite abelian group $G$ with the properties that A) the order of every element is a factor of $\sqrt{q}+1$ and B) the order meets the Hasse-Weil bound $\left\vert \#G-(q+1)\right\vert\le 2\sqrt q$.
