[Axiom-developer] [Q] How to classify radicalSolve(z^(1/1)=1) -> "There are..." ?

[Vladimir Bondarenko]

Good Day,

Should the behaviour (C) be interpreted as a bug or as a feature?

.......................................................................

(A) -> radicalSolve(z^1=1)

       [z= 1]

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(B) -> solve(z^1=1)

       [z= 1]


    -> solve(z^(1/1)=1)

       [z= 1]

.......................................................................
       
but

.......................................................................

(C) -> radicalSolve(z^(1/1)=1)

   There are 4 exposed and 0 unexposed library operations named
      radicalSolve having 1 argument(s) but none was determined to be
      applicable. Use HyperDoc Browse, or issue
                          )display op radicalSolve
      to learn more about the available operations. Perhaps
      package-calling the operation or using coercions on the arguments
      will allow you to apply the operation.

   Cannot find a definition or applicable library operation named
      radicalSolve with argument type(s)
                         Equation Expression Integer

      Perhaps you should use "@" to indicate the required return type,
      or "$" to specify which version of the function you need.


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The same with

radicalSolve(z^(2/2)=1)
radicalSolve(z^(3/3)=1)
radicalSolve(z^(4/4)=1)
radicalSolve(z^(5/5)=1)  etc
      

Much thanks in advance for the help.


Best wishes,

Vladimir Bondarenko