Taking cubes in Z/(17) turns out to be a bijection on (Z/(17))* := {1,..., 16}, i.e.,...

 

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Taking cubes in Z/(17) turns out to be a bijection on (Z/(17))* := {1,..., 16}, i.e., the map induced by x + x mod 17 is invertible. It turns out that cubing can in fact be inverted by taking another power, i.e., there is an e {1,...,16} such that (x) = =x mod 17, for all all x{1,..., 16}. Please find this magical e. Hint: Use Fermat's Little theorem. Taking cubes in Z/(17) turns out to be a bijection on (Z/(17))* := {1,..., 16}, i.e., the map induced by x + x mod 17 is invertible. It turns out that cubing can in fact be inverted by taking another power, i.e., there is an e {1,...,16} such that (x) = =x mod 17, for all all x{1,..., 16}. Please find this magical e. Hint: Use Fermat's Little theorem.