Show that 3 is a primitive squareroot mod 14 Then write the

Show that 3 is a primitive squareroot mod 14. Then, write the other primitive squareroots mod 14 in terms of powers of 3; how many are there?

Solution

Computing powers of 3 and modulo 14 gives

3=                                   3 mod 14

3^2=                                          9 mod 14

3^3=27=                                  13 mod 14

3^4=81=          11 mod 14

3^5=3^4*3=11*3=33               5 mod 14

3^6=3^5*3=5*3=                     1 mod 14

So we see powers of 3 generate all natural numbers less than 14 coprime to 14. Hence, 3 is a primitive root mod 14

3^k will not be a primitive root mod 14 if k is even because then only

3^{2m} will be generated ie even powers of 3 so only: 9,11 and 1 are generated

So we are left with:

3^3 and 3^5

Taking powers of 3^3 gives the powers

3,6,9,12,15,18 or

3,6,9,12,1,4 mod 14

But 3^6=1 So :3^{6+k}=3^k

So powers are:

3,6,3,6,1,4

Hence 3^3 is not a primitive root mod 14

Taking powers of 3^5 gives the powers

5,10,15,20,25,30or

5,10,1,6,11,2 mod 14

But 3^6=1 So :3^{6+k}=3^k

So powers are:

5,4,1,6,5,2

Hence 3^5 is not a primitive root mod 14