Suppose that a has order 3 modulo some p. Prove that a+1 has order 6 modulo p. [Hint: Take a^3 - 1 (which is 0 modulo p) and factor it to get a new expression for a+1.]
Well clearly $\displaystyle a \neq 1$. Since $\displaystyle 0=a^3-1=(a-1)(a^2+a+1)$ we have $\displaystyle 0=a^2+a+1$. I think you can take it from here!
Suppose that a has order 3 modulo some p. Prove that a+1 has order 6 modulo p. [Hint: Take a^3 - 1 (which is 0 modulo p) and factor it to get a new expression for a+1.]