Order of a polynomial over a finite field I want to show that the order of a polynomial $x^4+ax^2+b$ is a divisor of $2(q^2-1)$ over a finite field with $q$ elements.
 A: Sorry about jotting down something in haste. In addition to the obvious missing assumption $b\neq0$ we actually need to assume that the zeros of the
quadratic polynomial $p(x)=x^2+ax+b$ are distinct for the claim to hold. More details follow.
The polynomial $p(x)$ has its zeros, call them $\alpha_1$ and $\alpha_2$, in the field $K=\Bbb{F}_{q^2}$, so they have order dividing $q^2-1=|K^*|$. 
If $\alpha_1\neq\alpha_2$ we thus have
$$x^{q^2-1}\equiv1\pmod{x-\alpha_1}\quad\text{and}\quad x^{q^2-1}\equiv1\pmod{x-\alpha_1}.
$$
By the Chinese Remainder Theorem this tells us that
$$
x^{q^2-1}\equiv1\pmod{p(x)},
$$
and consequently (substitute $x^2$ for $x$ everywhere)
$$
x^{2(q^2-1)}\equiv1\pmod{x^4+ax^2+b}.
$$

The case $\alpha_1=\alpha_2$ is a bit different, because CRT doesn't bite as above when $x-\alpha_1$ and $x-\alpha_2$ are no longer coprime. It does follow that we then have $\alpha_1\in\Bbb{F}_q$, but the claim may actually be false in this case. The smallest counterexample is the case $p(x)=x^2+1=(x+1)^2$ when $q=2$. The original biquadratic is thus $x^4+1$, but this is obviously not a factor of $x^{2(q^2-1)}-1=x^6-1$. In this case (still assuming $b\neq0$) we have
$$
x-\alpha_1\mid x^{q-1}-1,
$$
so
$$
p(x)\mid (x^{q-1}-1)^2\mid (x^{q-1}-1)^p=x^{p(q-1)}-1,
$$
where $p$ is the characteristic. Again, plugging in $x^2$ for $x$ everywhere shows that the order of the quartic is a factor of $2p(q-1)$. Therefore the order is not necessarily a factor of $2(q^2-1)$. Another counterexample is the case $p(x)=(x+1)^2=x^2+2x+1$. It is easy to see that the order of the quartic $x^4+2x^2+1$ is $12\nmid 2(3^2-1)=16$.
