Remainder when $\prod_{n=1}^{100}(1- n^{2} +n^{4})$ is divided by $101$ 
What is the remainder when the expression
$$\prod_{n=1}^{100}(1- n^{2} +n^{4})$$
is divided by $101$?

If $\zeta=\dfrac{-1+\sqrt{-3}}{2}$, then
$$1-n^2+n^4=(1-n+n^2)(1+n+n^2)=(-\zeta-n)(-\bar{\zeta}-n)(\zeta-n)(\bar{\zeta}-n).$$
We then have
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv \prod_{n=1}^{100}\big((-\zeta-n)(-\bar{\zeta}-n)(\zeta-n)(\bar{\zeta}-n)\big)\pmod{101}\,.$$
Since
$$\prod_{n=1}^{100}(x-n)\equiv x^{100}-1\pmod{101},$$
we obtain
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv\big((-\zeta)^{100}-1\big)\big((-\bar\zeta)^{100}-1\big)\big(\zeta^{100}-1\big)\big(\bar{\zeta}^{100}-1\big)\pmod{101}\,.$$
Since $\zeta^3=1$ and $\bar{\zeta}^3=1$, we get
$$(-\zeta)^{100}=\zeta^{100}=\zeta\text{ and }(-\bar\zeta)^{100}=\bar\zeta^{100}=\bar\zeta\,.$$
Therefore,
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv (\zeta-1)^2(\bar{\zeta}-1)^2=\big((1-\zeta)(1-\bar{\zeta})\big)^2\pmod{101}\,.$$
As
$$(x-\zeta)(x-\bar{\zeta})=x^2+x+1\,,$$
we get
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv (1^2+1+1)^2=9\pmod{101}\,.$$
Are there other solutions?  How do we solve this problem without resorting to complex numbers?

Modulo $101$, the set of values $0^3, 1^3,\dots,100^3$ is a permutation of $0,1,2,\dots,100.$ This is because $101$ is prime and $3$ is not a divisor of $100.$
But $$n^4-n^2+1=\frac{n^6+1}{n^2+1}$$
Now, if $n=10,91$ then $n^2+1$ is divisible by $101.$ The other terms are a permutation, so:
$$\begin{align}\prod_{n=1}^{100} (n^4-n^2+1)&=(10^4-10^2+1)(91^4-91+1)\prod_{n\neq 10,91}\frac{n^6+1}{n^2+1}\\
&\equiv (10^4-10^2+1)((-10)^4-(-10)^2+1)\pmod{101}\\
&\equiv 3\cdot 3=9\pmod{101}
\end{align}$$

This works more generally if $p\equiv 5\pmod {12}:$
$$\prod_{n=1}^{p-1}\left(n^4-n^2+1\right)\equiv 9\pmod p$$
If $p\equiv 11\pmod{12},$ the remainder is $1.$
I think when $p\equiv 1\pmod{12},$ the remainder is $0.$
Not sure about $p\equiv 7\pmod{12}.$
