Can someone please help me with this one? N = Natural number, Z= integers.
Prove by induction that if n is a N and a,b is a Z and a congruent b(mod n), then a^m congruent b^m (mod n).
what I did was start with a + b = nk
then a = nk - b and use this fact in a^m -b^m = nk by rewriting it as aa^(m-1) - bb^(m-1) = 2k by subbing a into it. This did not work after multiple attempts. I am quite frustrated.