I collect the property of the solution for a1p-1+a2p-1+....+anp-1=bp-1.

 Proposition

       If there is a primitive solution to a1p-1+a2p-1+....+anp-1=bp-1,
       then b is not divisible by p and only one of (a1,...,an) is not divisible by p
       and others are divisible by p.

       Condition

              1.p   : odd prime
              2.n   : n<=p-1
              3.(a1,...,an),b: integer
              4.(a1,...,an,b) = 1
          
   
      Proof.
                                                                                      
          a1p-1+a2p-1+....+anp-1=bp-1..................(1)

          First,at (1) by Fermat's theorem(ap-1= 0,1 mod p),if b = 0 mod p,then all of (a1,...,an) = 0 mod p.
          But,this is not the primitive solution.
          
          So,b is not divisible by p because we are finding the primitive solution.

          When b <> 0 mod p then bp-1 = 1 mod p,so only one of (a1,...,an) = 1 mod p 
          and others of (a1,...,an) = 0 mod p.



 Example


       Case p=5,n=3

                   27676244+13904004+6738654 = 28130014

                  83322084+55078804+17055754 = 87074814

                 112890404+82825434+58700004 = 121974574

                141737204+125522004+44790314 = 160030174

                 162810094+70286004+36428404 = 164305134

                187967604+153656394+26824404 = 206156734

                      4145604+2175194+958004 = 4224814

       Case p=5,n=4


                  2724 +   3154 +   1204 +    304  = 3534
                  5994 +   4304 +   3404 +   2404  = 6514
                 13844 +   4354 +  24204 +   7104  = 24874
                 14324 +  23654 +  11904 +  11304  = 25014
                 15464 +  27454 +  10104 +   8504  = 28294
                 31524 +  23454 +  24604 +  22704  = 37234
                 16524 +  33954 +  32304 +   3504  = 39734 




       Case p=7

            No solutions are known to the 6.1.6 equation.



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