Euler Factorization For The Fifth Fermat number
Published On :2021-06-14 20:00:00
[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}]
