# $\frac{4^p - 1}{3}$ 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 $$n$$ divides $$2^{2p}-1$$.
Furthermore, $$2p$$ divides $$2\times\dfrac{(2^{p-1}-1)}3\times{(2^{p}+2)}=\dfrac{2^{2p}-4}3=n-1.$$
Therefore, $$n$$ divides $$2^{n-1}-1$$.