# primitive roots of 13

Given that 3 is a primitive root of 113, find 5 other primitive roots. 3×11 = 33 ≡ 2 [Hint: Recall that 2 is a primitive root modulo 13. We first want to find five positive integers that are relatively prime to 112. Let’s write this out. A primitive root of a prime p is an integer g such that g (mod p) has multiplicative order p-1 (Ribenboim 1996, p. 22). Least Primitive Roots Chapter 9 deals with estimates of the least primitive roots g(p) modulo p, a large prime. In fact, I have shown that g^11 is a primitive root mod 13. The topics investigated are listed below. One of the estimate here seems to A generator of this cyclic team is spoke of as a primitive root modulo m, or a primitive element of Zm×. To ﬁnd the other prinitive roots, use the table that was written down today in class.] We will choose the primes 5, 11, 13, 17, and 19, since all of them are relatively prime to 112. Their product 970377408 ≡ 1 (mod 31) and their sum 123 ≡ –1 (mod 31). The primitive roots are 3, 11, 12, 13, 17 ≡ –14, 21 ≡ –10, 22 ≡ –9, and 24 ≡ –7. The factors of m – 1 = 12 are 3, 2, and 2. Multiplicative team of integers modulo m, this team is cyclic if and provided that m is equivalent[a million] to a million, 2, 4, pk, or 2 pk the place pk is a capability of a wierd top extensive type. Evan Chen 3 Primitive Roots Example 3.3 (Primitive Roots Modulo 11 and 13) It turns out that g= 2 is a primitive root modulo both 11 and 13. But my question is how can I use this information to deduce that the product of all the primitive roots mod 13 is congruent to 1 mod 13. More generally, if GCD(g,n)=1 (g and n are relatively prime) and g is of multiplicative order phi(n) modulo n where phi(n) is the totient function, then g is a primitive root of n (Burton 1989, p. 187). I'm aware of the condition for k to such that g^k is a primitive root mod 13. 1.1. A few topics in the theory of primitive roots modulo primes p≥ 2, and primitive roots modulo integers n≥ 2, are studied in this monograph. Hence, a = 2 is the smallest primitive root. Hence, the primitive roots for m = 13 are 2, 2 5 mod 13 = 6, 2 7 mod 13 = 11, and 2 11 mod 13 = 7. (1) Find the index of 5 relative to each of the primitive roots of 13. Then the values of n < m – 1 that have no factors (except 1) in common with 3 and 2 are n = 5, 7, and 11. Since primitive roots are of the form 3 i where gcd(i, φ (17)) = 1, the primitive roots are 3, 10, 11, 14, 7, 12, 6, 5 (d) We showed above that the primitive roots of 18 are 5 and 11.

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