Euler Factorization For The Fifth Fermat number

 

$$\begin{array}{l}{F}_5=2^{2^5}+1=4294967297\\ \\ TheFermatnumber{F}_{5}isdivisibleby641.\\ \\ Proof:\\ \\ F_5=2^{32}+1=2^{18}\ast 2^{14}+1\\ \\ =262144\ast 2^{14}+1=261735\ast 2^{14}+409\ast 2^{14}+1\\ \\ =261735\ast 2^{14}+52352\ast 2^7+1\\ \\ =5\ast 52347\ast 2^{14}+\left(5+52347\right)\ast 2^7+1\\ \\ =\left(5\ast 2^7+1\right)\left(52347\ast 2^7+1\right)=641\ast 6700417.\\ \\ Whichgives641|F_5\end{array}$$