\[\begin{array}{l}
{F_5} = {2^{{2^5}}} + 1 = 4294967297\\\\
The{\rm{ }}Fermat{\rm{ }}number\;{F_5}\;is{\rm{ }}divisible{\rm{ }}by{\rm{ }}641.\\\\
Proof:\\\\
{F_5} = {2^{32}} + 1 = {2^{18}} * {2^{14}} + 1\\\\
= 262144 * {2^{14}} + 1 = 261735 * {2^{14}} + 409 * {2^{14}} + 1\\\\
= 261735 * {2^{14}} + 52352 * {2^7} + 1\\\\
= 5 * 52347 * {2^{14}} + \left( {5 + 52347} \right) * {2^7} + 1\\\\
= \left( {5 * {2^7} + 1} \right)\left( {52347 * {2^7} + 1} \right) = 641 * 6700417.\\\\
Which{\rm{ }}gives\;641\;|\;{F_5}\\
\end{array}\]
