$\frac{4^p - 1}{3}$ is a Fermat pseudoprime with respect to 2 I have to prove that $n = \frac{4^p - 1}{3}$ is a Fermat pseudoprime with respect to $2$ when $p \geq 5$ is a prime number. I have proved that $n$ is not prime because $4^p - 1 = (2^p-1)(2^p+1)$ and $(2^p + 1)$ is divisible by $ 3$. But now I can't show that $2^{n-1} \equiv 1\bmod n$.
I calculated that $2^{n-1} = 2^{(2^p + 2)(2^p-2)/3}$ but I don't know if I can deduce anything from this.

$n=\dfrac{4^p-1}3=\dfrac{2^{2p}-1}3,$ so $n\mid 2^{2p}-1,\,$ so $\,\color{#c00}{2^{2p}\equiv 1}\pmod{\!n}$
Further, $2p$ divides $2\times\dfrac{(2^{p-1}-1)}3\times{(2^{p}+2)}=\dfrac{2^{2p}-4}3=n-1,\,$ so $n-1 = 2pk$
Therefore, $\!\bmod n,\,$ we have $\,2^{n-1}\equiv (\color{#c00}{2^{2p}})^{k}\equiv \color{#c00}1^k\equiv 1$
