Solution
= Solution
Since $p\ge5$ is odd, $2^p\equiv2\pmod3$ and
$$
N=\frac{4^p-1}{3}=(2^p-1)\frac{2^p+1}{3}
$$
is a product of two odd <integers> greater than one. Thus $N$ is odd and composite. Also $2^{2p}=4^p\equiv1\pmod N$. Fermat's theorem modulo $p$ gives $p\mid4^{p-1}-1$, and $p\ne3$, so $p\mid N-1=4(4^{p-1}-1)/3$. This number is even; hence $2p\mid N-1$. Therefore
$$
\boxed{2^{N-1}\equiv1\pmod N,}
$$
proving that $N$ is a base-two <Fermat pseudoprime>.