Show that if a^k≡ b^k (mod m) and a^(k+1) ≡ b^(k+1) (mod m), where a,b,k, and m are integers with k > 0, m > 0 such that (a,m) = 1, then a ≡ b (mod m). If the condition (a,m) is dropped, is the conclusion that a ≡ b still valid?

just throwing out my brain storm ideas, but is this an induction problem, or do i have to set k=1 or something else?..