Let a b m Z Suppose a b m Use the result of Exercise 26b in

Let a, b, m Z. Suppose a = b m. Use the result of Exercise 26(b) in Section 4 to prove by induction that for each n N, an 6n mod m. Let a, b, m Z and let n N. Use the result of Exercise 7 to prove that if m divides then m divides b-a, then m divides bn - an.

Solution

Assume n divides a-b, then a=nx+b. Therefore a2=(nx+b)2=n2x2+2bnx+b2=n(nx2+2bx)+b2. Since a2=n(nx2+2bx)+b2 and nx2+2bx is an integer, a2-b2=n(nx2+2bx). Therefore a2b2(mod n).

You have already stated that a-b=nx, or a=nx+b, for some integer x.