Show that $F_5 = 2^{2^5}$ + 1 (Fermat number with $n = 5$) is divisible by 641.
We have
$$ \begin{align} 2^{2^5} & = 2^{32} + 1= (2 ^ 8)^{4} + 1 \\ \\ & \equiv 256^4 + 1 \pmod{641} \\ \\ & \equiv 640 + 1 \pmod{641} \\ \\ & \equiv (-1) + 1\pmod {641} \\ \\ & \equiv 0 \pmod{641} \end{align} $$
We use the following result in the above
$ 2^8 = 256 \equiv 256 \pmod{641} $
$ 256^4 = 4294967296 \equiv 640 \pmod{641} $
This means $F_5$ is divisible by 641, and is not a prime.