When p 2 x 1 is a solution since 1 1 1 mod 2 Suppose that p 1 mod 4 Then 4 p 1

# When p 2 x 1 is a solution since 1 1 1 mod 2 suppose

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When p = 2, x = 1 is a solution since 1 1 + 1 0 mod 2 . Suppose that p 1 mod 4. Then 4 | ( p - 1), so by the Primitive Root Theorem, there is an integer a such that ord p ( a ) = 4 . Thus, a 4 1 mod p, so a 4 - 1 ( a 2 - 1)( a 2 + 1) 0 mod p. This implies that either a 2 - 1 0 mod p or a 2 + 1 0 mod p. Note that a 2 - 1 0 mod p since ord p ( a ) = 4. Thus, a 2 + 1 0 mod p, so a is a solution of x 2 + 1 0 mod p. Suppose that p 3 mod 4. Then 4 | ( p - 3), so 4 ( p - 1). Since p is odd, p - 1 is even, so 2 | ( p - 1). Thus, ( p - 1 , 4) = 2. Now, suppose (by way of contradiction) that there is an integer a such that a 2 + 1 0 mod p. Then a 4 ( a 2 ) 2 ( - 1) 2 1 mod p. By Fermat’s Little Theorem, we also know that a p - 1 1 mod p. Since 2 = ( p - 1 , 4), there are integers u and v such that 2 = ( p - 1) s + 4 t . Thus, we have a 2 a ( p - 1) s +4 t ( a p - 1 ) s ( a 4 ) t 1 mod p. Then a 2 1 ≡ - 1 mod p, 87
Stanford University EPGY Number Theory so 2 0 mod p. This implies that p | 2 , which is a contradiction since p is an odd prime. Thus, there is no integer a such that a 2 + 1 0 mod p , so the congruence x 2 + 1 0 mod p does not have any solutions in this case. Finally, we note that the ideas of order and primitive roots can be extended to all (i.e. not necessarily prime) positive integers. Definition 12.3 Suppose that ( a, n ) = 1. The order of a modulo n is the smallest positive integer b such that a b 1 mod n. By the Euler-Fermat Theorem, we know that a φ ( n ) 1 mod n. Thus, ord n ( a ) φ ( n ). Using an argument similar to that given in the proof of Theorem 12.1, we can prove the following result. Theorem 12.5 Suppose that ( a, n ) = 1 and that a b 1 mod n. Then ord n ( a ) | b, and in particular, ord n ( a ) | φ ( n ) . Definition 12.4 If ( a, n ) = 1 and ord n ( a ) = φ ( n ), then we say that a is a primitive root modulo p . Example 12.6 3 is a primitive root modulo 10 since φ (10) = 4 and 3 1 3 mod 10, 3 2 9 mod 10, 3 3 7 mod 10, 3 4 1 mod 10. 88
Stanford University EPGY Number Theory Problem Set 1. Compute each of the following. (a) ord 5 (3) (b) ord 5 (4) (c) ord 7 (3) (d) ord 9 (2) (e) ord 15 (2) (f) ord 16 (3) (g) ord 10 (3) 2. Find all primitive roots modulo 5. 3. Find all primitive roots modulo 7. 4. (a) Find all primitive roots modulo 13. (b) For each number d dividing 12, list the a ’s such that 1 a < 13 and ord 13 (1) = d . 5. Find all primes less than 20 for which 3 is a primitive root. 6. In this exercise, you will investigate the value of ordn(2) for odd integersn. (a) Compute the value of ord n (2) for each odd number 3 n 19. (b) In the table below, the value of ord n (2) is given for all odd numbers between 21 and 115. ord 21 (2) = 6 ord 23 (2) = 11 ord 25 (2) = 20 ord 27 (2) = 18 ord 29 (2) = 28 ord 31 (2) = 5 ord 33 (2) = 10 ord 35 (2) = 12 ord 37 (2) = 36 ord 39 (2) = 12 ord 41 (2) = 20 ord 43 (2) = 14 ord 45 (2) = 12 ord 47 (2) = 23 ord 49 (2) = 21 ord 51 (2) = 8 ord 53 (2) = 52 ord 55 (2) = 20 ord 57 (2) = 18 ord 59 (2) = 58 ord 61 (2) = 60 ord 63 (2) = 6 ord 65 (2) = 12 ord 67 (2) = 66 ord 69 (2) = 22 ord 71 (2) = 35 ord 73 (2) = 9 ord 75 (2) = 20 ord 77 (2) = 30 ord 79 (2) = 39 ord 81 (2) = 54 ord 83 (2) = 82 ord 85 (2) = 8 ord 87 (2) = 28 ord 89 (2) = 11 ord 91 (2) = 12 ord 93 (2) = 10 ord 95 (2) = 36 ord 97 (2) = 48 ord 99 (2) = 30 ord 101 (2) = 100 ord 103 (2) = 51 ord 105 (2) = 12 ord 107 (2) = 106 ord 109 (2) = 36 ord 111 (2) = 36 ord 113 (2) = 28 ord 115 (2) = 44

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• Fall '16
• jena smith
• Number Theory, Prime number, Stanford University EPGY