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By Pierre Deligne

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The conjecture is still unproved. See L. J. Goldstein [1971] for details on the state of the conjecture twenty years ago. Other references are Shanks [1985] and Silver-man [1997]. Artin actually had a more specific conjecture to the effect that if a =£ — 1 and a / square, then approximately 3/8 of all primes will have a for a primitive root. The heuristic probabilistic argument that led to this conjecture will be discussed soon. The conjecture needs modification for certain values of a. 6). 6. A computer verification ofArtin 's conjecture by Baillie a # Primes < 33 x 106 with a as primitive root Artin factor, A in (6) 759,733 759,658 759,754 759,754 800,218 20 799,741 = —759,754 6 7 760,037 760,133 759,754 759,754 8 455,894 455,854 = ^759,754 10 11 12 760,192 760,352 759,988 759,754 759,754 759,754 13 764,719 764,655 = 11759,754 by (6) below, then, since the number of primes < 33 x 106 is 2,031,667, we have A-2,031,667 = 759,754.

Exercise. Show that the sum of all the primitive roots mod p is congruent to li(p — 1) mod p. , are distinct primes. Exercise. Show that if n > 1 and /x denotes the Mobius function of the preceding exercise, we have d\n d>0 Exercise. The Mobius Inversion Formula. Suppose that / ; Z + —> C and let ix denote the Mobius function defined in the exercise above. Let F(n) = ^ f(d). d\n d>0 Show that d\n d>0 A Few Remarks on Multiplicative Functions We have given three examples of multiplicative functions / : Z + -> Z + in this section: 0(«), ok{n), and /x(n).

By the Chinese remainder theorem, (Z/12Z)* = (Z/3Z)* 0 (Z/4Z)*. So every element of (Z/12Z)* has order 2. So there can be no primitive root modulo 12. That is, (Z/12Z)* is not a cyclic group. Exercise. Write a Mathematica program to find a primitive root mod n if it exists. Note. Although we can prove that primitive roots exist for any prime modulus, we won't be able to provide an easy way to find them. Trial and error is the standard method. That would be rather time consuming for large moduli.

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