Galois discovered Quintic Equation has no radical (expressed with +,-,*,/, nth root) solutions, but his new Math “Group Theory” also explains:

$latex x^{5} – 1 = 0 text { has radical solution}$

but

$latex x^{5} -x -1 = 0 text{ has no radical solution}$

Why ?

$latex x^{5} – 1 = 0 text { has 5 solutions: } displaystyle x = e^{frac{ikpi}{5}}$

$latex text{where k } in {0,1,2,3,4}$

which can be expressed in

$latex x= cos frac{kpi}{5} + i.sin frac{kpi}{5} $

hence in {+,-,*,/, √ }

ie

$latex x_0 = e^{frac{i.0pi}{5}}=1$

$latex x_1 = e^{frac{ipi}{5}}$

$latex x_2 = e^{frac{2ipi}{5}}$

$latex x_3 = e^{frac{3ipi}{5}}$

$latex x_4 = e^{frac{4ipi}{5}}$

$latex x_5 = e^{frac{5ipi}{5}}=1=x_0$

=>

$latex text {Permutation of solutions }{x_j} text { forms a Cyclic Group: }

{x_0,x_1,x_2,x_3,x_4} $

**Theorem: All Cyclic Groups are Solvable **

=>

$latex x^{5} -1 = 0 text { has radical solutions.}$

However,

$latex x^{5} -x -1 =…

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