TITLE: Primality test involving quartic fields and polynomials
QUESTION [6 upvotes]: Let $n$ be an odd number, $D=a^2+4$ and $(D | n)=-1$ where $(D | n)$ is the Jacobi symbol. 
Then $n$ is prime if and only if
$(x^2 + ax)^n = x^3 - x^2 + (D-2)x - D \pmod {{x^4 + Dx^2 + D},n}$ 
or
$(x^2 - ax)^n = x^3 - x^2 + (D-2)x - D \pmod {{x^4 + Dx^2 + D},n}$
I have constructed a proof (here) for the converse of this test but not the actual test itself. Are there counterexamples? A heuristic argument suggesting a counterexample? 
There are no counterexamples for any $(|a|,n) < 10000$.
The idea for this algorithm came from a version of Agrawal's Conjecture with $r=5$ involving a quartic polynomial, without resorting to higher-degree polynomials (larger values of $r$). It is much easier to use a constructed family of cyclic quartic polynomials with one free parameter.
You can run the test here;
Also related is this conjecture/test here (which also gave me an idea for this test).

REPLY [2 votes]: This is a partial answer.
This answer proves the following claim :
Claim : If $n$ is an odd prime, then 
$$(x^2 \pm  ax)^n\equiv \pm ab_nx^3-x^2\mp aDb_{n-2}x-D\pmod{x^4 + Dx^2 + D,n}$$
where$$b_n=\frac{(-D+|a|\sqrt D)^{(n-1)/2}-(-D-|a|\sqrt D)^{(n-1)/2}}{2^{(n-1)/2}|a|\sqrt D}$$

The claim follows from the following four lemmas :
Lemma 1 : $$(x^2 \pm ax)^n\equiv \pm ax^n+x^{2n}\pmod{n}$$
Lemma 2 : $$\pm ax^n+x^{2n}\equiv \pm ab_nx^3+b_{2n+1}x^2\mp aDb_{n-2}x-Db_{2n-1}\pmod{x^4 + Dx^2 + D}$$where$$b_n=\frac{(-D+|a|\sqrt D)^{(n-1)/2}-(-D-|a|\sqrt D)^{(n-1)/2}}{2^{(n-1)/2}|a|\sqrt D}$$
Lemma 3 : $$b_{2n+1}\equiv -1\pmod n$$
Lemma 4 : $$Db_{2n-1}\equiv D\pmod n$$

Lemma 1 : $$(x^2 \pm ax)^n\equiv \pm ax^n+x^{2n}\pmod{n}$$
Proof for lemma 1 :
By the binomial theorem, we have
$$(x^2\pm ax)^n=\sum_{k=0}^{n}\binom nk(x^2)^{n-k}(\pm ax)^k$$
Since $\binom nm\equiv 0\pmod n$ for $1\le m\le n-1$, we get
$$(x^2\pm ax)^n\equiv (\pm ax)^n+x^{2n}=\pm a^nx^n+x^{2n}\equiv \pm ax^n+x^{2n}\pmod n$$

Lemma 2 : $$\pm ax^n+x^{2n}\equiv \pm ab_nx^3+b_{2n+1}x^2\mp aDb_{n-2}x-Db_{2n-1}\pmod{x^4 + Dx^2 + D}$$
Proof for lemma 2 : 
Let $b_nx^3+c_nx^2+d_nx+e_n$ be the remainder when $x^n$ is divided by $x^4 + Dx^2 + D$. Then, we get
$$b_{n+1}=c_n,\quad c_{n+1}=-Db_n+d_n,\quad d_{n+1}=e_n,\quad e_{n+1}=-Db_n$$
from which we have
$$b_{n+2}=-Db_{n}-Db_{n-2}$$
which can be written as
$$b_{n+2}+kb_n=m(b_{n}+kb_{n-2})$$
where $$m=\frac{-D+|a|\sqrt D}{2},\qquad k=\frac{D+|a|\sqrt D}{2}$$
So, we get
$$b_n+kb_{n-2}=m(b_{n-2}+kb_{n-4})=\cdots =m^{(n-3)/2}(b_{3}+kb_1)=m^{(n-3)/2}$$
Dividing by $(i\sqrt k)^n$ gives
$$f_n-f_{n-2}=\frac{m^{(n-3)/2}}{(i\sqrt k)^n}:=g(n)$$
where $f_n=\frac{b_n}{(i\sqrt  k)^n}$.
So, we get
$$f_n=f_1+\sum_{j=1}^{(n-1)/2}g(2j+1)$$
from which we have
$$b_n=(i\sqrt k)^nf_n=\frac{(-D+|a|\sqrt D)^{(n-1)/2}-(-D-|a|\sqrt D)^{(n-1)/2}}{2^{(n-1)/2}|a|\sqrt D}$$
Finally, we get
$$\begin{align}\pm ax^n+x^{2n}&\equiv \pm a(b_nx^3+c_nx^2+d_nx+e_n)+b_{2n}x^3+c_{2n}x^2+d_{2n}x+e_{2n}
\\\\&\equiv \pm ab_nx^3+b_{2n+1}x^2\mp aDb_{n-2}x-Db_{2n-1}\pmod{x^4 + Dx^2 + D}\end{align}$$

Lemma 3 : $$b_{2n+1}\equiv -1\pmod n$$
Proof for lemma 3 : 
From lemma 2, we get
$$\begin{align}2^{n-1}b_{2n+1}&=\frac{(D+|a|\sqrt D)^{n}+(-D+|a|\sqrt D)^{n}}{2|a|\sqrt D}
\\\\&=\frac{1}{2|a|\sqrt D}\sum_{k=0}^{n}\binom nk(|a|\sqrt D)^{n-k}(D^k+(-D)^k)
\\\\&=\frac{1}{|a|\sqrt D}\sum_{j=0}^{(n-1)/2}\binom n{2j}(|a|\sqrt D)^{n-2j}D^{2j}
\\\\&=\sum_{j=0}^{(n-1)/2}\binom n{2j}(|a|\sqrt D)^{n-2j-1}D^{2j}
\\\\&\equiv a^{n-1}\cdot D^{(n-1)/2}\pmod n
\\\\&\equiv -1\pmod n\end{align}$$
from which we have
$$b_{2n+1}\equiv -1\pmod n$$

Lemma 4 : $$Db_{2n-1}\equiv D\pmod n$$
Proof for lemma 4 : 
$$\begin{align}2^{n}b_{2n+3}&=-\frac{1}{2|a|\sqrt D}(D+|a|\sqrt D)(D+|a|\sqrt D)^{n}
\\&\qquad\qquad +\frac{1}{2|a|\sqrt D}(-D+|a|\sqrt D)(-D+|a|\sqrt D)^{n}
\\\\&=-\frac{\sqrt D}{2|a|}\bigg((D+|a|\sqrt D)^{n}+(-D+|a|\sqrt D)^{n}\bigg)
\\&\qquad\qquad +\frac 12\bigg(-(D+|a|\sqrt D)^{n}+(-D+|a|\sqrt D)^{n}\bigg)
\\\\&=-\frac{\sqrt D}{2|a|}\sum_{k=0}^{n}\binom nk(|a|\sqrt D)^{n-k}(D^k+(-D)^k)
\\&\qquad\qquad +\frac 12\sum_{k=0}^{n}\binom nk(|a|\sqrt D)^{n-k}(-D^k+(-D)^k)
\\\\&=-\sum_{j=0}^{(n-1)/2}\binom n{2j}(|a|\sqrt D)^{n-2j-1}D^{2j+1}
\\&\qquad\qquad-\sum_{j=1}^{(n+1)/2}\binom n{2j-1}(|a|\sqrt D)^{n-2j+1}D^{2j-1}
\\\\&\equiv -a^{n-1}D^{(n+1)/2}-D^n\pmod n
\\\\&\equiv 0\pmod n\end{align}$$
from which we have
$$b_{2n+3}\equiv 0\pmod n$$
So, we get
$$Db_{2n-1}\equiv -b_{2n+3}-Db_{2n+1}\equiv D\pmod n$$