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If p | n, then p \ n - 1, and then for any g E G, g ± 1, we have gn~l ^ 1 (mod n). 64, the sets S and gS are disjoint. 65, the sets gS, g E G) are disjoint. Then [U^G Sg\ = \G\ • \S\ = p|5|, and p\S\ < |(Z/nZ)*| = >(n), and therefore \s\ < ^ < k z /n Z ) * |, p 4 since p > 5 by the assumption of the theorem. 67. 68. Let n = P1P2, where p\ and p2 are distinct primes. Then n - 1 is not divisible by one of the two numbers Pi — 1. 69. Let n = P1P2, where p\ ^P2- Then |5| < (p(n)/4. 8. MODERN METHODS FOR PRIMALITY TESTING 23 PROOF.
Suppose A'(n) | n —1 andn is of type A. If a g N, 1 < a < n, and ( |) = —1, then either (a,n) ^ 1,n or, for some k, 1 < k < 1/2(n — 1), g c d ^ a ^ - 1) (mod n ),n j ^ 1,n. Thus if, in Miller’s algorithm, we reach this value of a, then in part (iii) of Step 2 we can conclude that n is composite. , the algorithm is correct for n of type A. 53. Suppose A'(n) | n - 1 andn is of type B. If a e N, 1 < a < n, and ( ^ ) = —1, then either (a,n) ^ l,n , or, for some k, 1 < k < ^ ( n —1), gcd^(a^~ - 1) (mod ri),nj ^ 1,n.
Then g(x)pd = g(x) (mod /i(x)) and therefore g(x)p _1 = 1(mod h(x)). This means that pk —1 | pd — 1, and therefore k | d. Now, x r = 1(mod h(x)) in Z/pZ[x]. Since xr — 1 has no multiple irreducible factors in Z/pZ[x], we have h(x) ^ x — 1. Therefore, the order of x (mod h(x)) is a prime number r. , pk = 1 (mod r). It now follows from the definition of d that d | k . It follows from the above argument that k = d, proving the last assertion of the lemma. □ The next two lemmas contain results about the distribution of primes.









