For such a prime modulus generator all primitive roots produce full cycles. Thus, first find a small primitive root, i.e., find an a such that the smallest integer k that satisfies a k mod 13 = 1 is k = m – 1 = 12. [Hint: Recall that 2 is a primitive root modulo 13. Primitive root of a prime number n is an integer r between[1, n-1] such that the values of r^x(mod n) where x is in range[0, n-2] are different. To find the other prinitive roots, use the table that was written down today in class.] Example 1. Use the fact: If g is a primitive root mod p, then one of g and g+p (whichever one is odd) is a primitive root mod 2p. (1) Find the index of 5 relative to each of the primitive roots of 13. 2) Next, we find a primitive root mod 82. 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. Then, 6 + 41 = 47 is a primitive root mod 82 (since 47 is odd). By (1), 6 is a primitive root mod 41. In fact, I have shown that g^11 is a primitive root mod 13. We see from the table that 2 is a primitive root mod 13… 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. If b is a primitive root mod 13, th en the complete set of primitive roots is {b 1, b 5, b 7, b 11}. The number of primitive roots mod p is ϕ (p − 1). I'm aware of the condition for k to such that g^k is a primitive root mod 13. Given that 2 is a primitive root of 59, find 17 other primitive roots of 59. We know that 3, 5, 7, 11, 13… (3) Find the number of roots mod 13 for each of the following polynomials: (a) x 2 + 3 x + 2 (b) x … It is easily verified that 2 k mod 13 = 2, 4, 8, 3, 6, 12, 11, 9, … For example, consider the case p = 13 in the table. ϕ (p − 1) = ϕ (12) = ϕ (2 2 3) = 12(1 − 1/2)(1 − 1/3) = 4. Suppose that we have a primitive root, g. For example, 2 is a primitive root of 59. Then it turns out for any integer relatively prime to 59-1, let's call it b, then $2^b (mod 59)$ is also a primitive root of 59.