Show that, for each $n \geq 2$, the whole number $4n^{2}-1$ is a Fermat pseudoprime to base $2n$

Question

I need to show that, for each $n \geq 2$, the whole number $4n^{2}-1$ is a Fermat pseudoprime to base $2n$; (hint: $4n^{2}-2$ is a multiple of $2$).

Fermat pseudoprime to base $2n$ means that $(2n)^{4n^{2}-2} \equiv 1 \mod 4n^{2}-1$, but from this point on I don't know what to do.

Thanks.

Answer 1

Another hint besides the one given:

$$(2n)^2\equiv 1\bmod 4n^2-1$$

Answer 2

$a$ is a Fermat psp base $m\iff\bmod m\!:\ a^{m-1}\equiv 1\iff {\rm ord}(a)\mid m\!-\!1,\,$ by the Order Theorem

Thus in OP we need: $\ {\rm ord}(2n)\mid 4n^{\large 2}-2 = \color{#c00}2(2n^{\large 2}-1)\,$ with an obvious $\rm\color{#c00}{small\ factor}$ to try first.

It works: $\!\bmod 4n^{\large 2}\!-1\!:\,\ 4n^{\large 2}\equiv 1\,\Rightarrow\, (2n)^{\large\color{#c00}2}\equiv\ \ldots $