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Download Symposium in Honor of C. H. Clemens by Bertram A., Carlson J.A., Kley H. (eds.) PDF

By Bertram A., Carlson J.A., Kley H. (eds.)

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Pr of p − 1. 3. ] If for every pi , we have a(p−1)/pi ≡ 1 (mod p), then a is a generator of (Z/pZ)∗ , so output a and terminate. 4. [Try next] Set a = a + 1 and go to Step 3. Proof. Let a ∈ (Z/pZ)∗ . The order of a is a divisor d of the order p − 1 of the group (Z/pZ)∗ . Write d = (p − 1)/n, for some divisor n of p − 1. If a is not a generator of (Z/pZ)∗ , then since n | (p − 1), there is a prime divisor pi of p − 1 such that pi | n. Then a(p−1)/pi = (a(p−1)/n )n/pi ≡ 1 (mod p). Conversely, if a is a generator, then a(p−1)/pi ≡ 1 (mod p) for any pi .

This implies that a | 1, which is a contradiction since a ≥ 2. Suppose n ∈ N. 13, we can either quickly prove that n is not prime, or convince ourselves that n is likely prime (but not quickly prove that n is prime). For example, if 2n−1 ≡ 1 (mod n), then we have proved that n is not prime. On the other hand, if an−1 ≡ 1 (mod n) for a few a, it “seems likely” that n is prime, and we loosely refer to such a number that seems prime for several bases as a pseudoprime. 4 Primality Testing 37 There are composite numbers n (called Carmichael numbers) with the amazing property that an−1 ≡ 1 (mod n) for all a with gcd(a, n) = 1.

5 The Structure of (Z/pZ)∗ This section is about the structure of the group (Z/pZ)∗ of units modulo a prime number p. The main result is that this group is always cyclic. We will use this result later in Chapter 4 in our proof of quadratic reciprocity. 40 2. 1 (Primitive root). A primitive root modulo an integer n is an element of (Z/nZ)∗ of order ϕ(n). We will prove that there is a primitive root modulo every prime p. Since the unit group (Z/pZ)∗ has order p−1, this implies that (Z/pZ)∗ is a cyclic group, a fact that will be extremely useful, since it completely determines the structure of (Z/pZ)∗ as a group.

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