I) For any integers a _p 0, r and s, if ra _p sa, then r _p s. (Note that you dont yet know that division modulo p exists.) [DONE]

II) For any integer a is not congruent to 0 mod p ( a _p 0) , the values a%p, 2a%p, 3a%p,...,(p1)a%p hit every element in the

set 1, 2, 3, . . . , p 1 exactly once. (Hint: use the previous part.)

III) Show that every non-zero element modulo p has an inverse: for every a 0, there exists b such that ab _p 1. Thus, division modulo p makes sense.