 Solution

Since $p\ge 5$ is odd, $2^p\equiv 2(\mathrm{mod}3)$ andis a product of two odd integers greater than one. Thus $N$ is odd and composite. Also $2^{2p}=4^p\equiv 1(\mathrm{mod}N)$. Fermat's theorem modulo $p$ gives $p\mid 4^{p-1}-1$, and $p\ne 3$, so $p\mid N-1=4(4^{p-1}-1)/3$. This number is even; hence $2p\mid N-1$. Thereforeproving that $N$ is a base-two Fermat pseudoprime.