Let d f and n be integers with nd1 also let AkZn fk0 Prove

Let d, f and n be integers with n>d>1 also let A={[k]Zn | [f][k]=[0]}. Prove if d=gcd(f, n), then the set A contains more than one element and fewer than n elements. [Note that Zn contains exactly n elements: [0], [1], …, [n-1].]

Solution

Given:Let d, f and n be integers with n>d>1

         and if d=gcd(f, n)

   From the relation: n>d>1 or 1<d<n

            that means d is between 1 and n.

   if d=gretest common division

      it may be 1 to n.

and set A={[k]Zn | [f][k]=[0]}.

         it is depend on the values in k, but

             this only contain these in Zn.

so, A contains more than one element and fewer than n elements.

  

Let d, f and n be integers with n>d>1 also let A={[k]Zn | [f][k]=[0]}. Prove if d=gcd(f, n), then the set A contains more than one element and fewer than

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