Assume p is a prime a and has order 5 and modulo p show that

Assume p is a prime a and has order 5 and modulo p, show that a^3 also has order 5 modulo p

a has order 5 modulo p means 5 is the smallest natural number ,k so that a^k=1 modulo p

(a^3)^5=(a^5)^3=1 modulo p

Hence order of a^3 is less than equal to 5

It cannot be 1 else a^3=1 which is a contradiction

Case 1. order =2

(a^3)^2=a^6=a=1

But a cannot be 1

Case 2. Order =3

(a^3)^3=a^9=a^4=1

THis is a contradiction

Case 3. Order =4

(a^3)^4=a^12=a^2=1

Again a contradiction.

HEnce, a^3 has order 5 modulo p

Case 4.