Show that if gcd(a, n) = gcd(a-1,n) = 1, then 1+a+a² + .... + a^(ϕ(n)-1) ≡ 0 (mod n). [Hint : Recall that a^(ϕ(n)) - 1 = (a-1) (a^(ϕ(n)-1) + a + ... + a² + a+1)].