Answer Happy

Let p be a prime and a EF, The set {D EN:ad = 1F,} contains p-1 by Fermat's little theorem, and in particular, is nonempty. We define the (mul- tiplicative) order of a, denoted |al, to be the smallest element of the displayed set (which exists by well-ordering). In other words, lal is the smallest positive integer with the property that alal 1F,. Note that, by Fermat’s little theorem, lal <p – 1 for all a EF. The orders of the nonzero elements in F7 can be read off from the table below: Q a a3 a a a6 1+7Z 1+7Z 1+7Z 1+7Z 1+7Z 1+7Z 2+7Z 4+7Z 1+7Z 2 +7Z 4+7Z 1+7Z 3+7Z 2 +7Z 6+7Z 4+7Z 5+7Z 1+7Z 4+7Z 2 + 7Z 1+7Z 4+ 7Z 2 + 7Z 1+7Z 5+7Z 4+7Z 6+7Z 2 + 7Z 3+7Z 1+7Z 6+7Z 1+7Z 6+7Z 1+7Z 6+7Z 1+7Z = p We see that |1 + 7Z| = 1, 12 + 7Z| = 3, 13+ 72|| 6, |4 + 7Z| = 3, 15 + 7Z = 6, and 16 + 72 = 2. 1. Let p be prime, a € Fm, and n e Z. Prove that an = 1f, if and only if |a| divides n (Hint: The “if statement is straightforward. For the "only if” statement, use long division to write n= |a|q+r with 0 <r<la). Prove that r = 0 by verifying that a" = 15, How does this imply the desired conclusion?) 2. Let p be a prime. Use the result of the previous problem to prove the following statements: (a) For every a € F, |a| divides p - 1. (b) If m and n are integers, then am=an if and only if m= n (mod al).