For any element a∈Zn∗ a \in {\mathbb Z}_n^* a∈Zn∗, consider the sequence of its powers 1,a,a2,a3,… 1,a,a^2,a^3,\ldots1,a,a2,a3,… mod n n n. All of the powers of a a a are in the finite set Zn∗ {\mathbb Z}_n^*Zn∗, so the sequence of powers of a a a repeats eventually. However, 32 2 mod 7;33 6 1 mod 7: Since the order of an element divides the order of the group, which is 6 in Mar 21, 2010 #1 Suppose q is a prime and p=4q+1 is a prime. University Math Help. General Wikidot.com documentation and help section. In fact, Euler's theorem implies that it repeats with period ϕ(n): \phi(n):ϕ(n): a0≡1≡aϕ(n)(modn).a^0 \equiv 1 \equiv a^{\phi(n)} \pmod n.a0≡1≡aϕ(n)(modn). Given that 2 is a primitive root of 59, find 17 other primitive … In general, since every nonzero element of Zp {\mathbb Z}_p Zp can be written as ga g^a ga mod p p p for some integer x x x, the equation xk≡y x^k \equiv y xk≡y mod p p p can be rewritten (ga)k≡gb, (g^a)^k \equiv g^b, (ga)k≡gb, or gak−b≡1 g^{ak-b} \equiv 1 gak−b≡1 mod p p p. Because g g g is a primitive root, this happens if and only if (p−1)∣(ak−b) (p-1)|(ak-b) (p−1)∣(ak−b), so ak≡b ak \equiv b ak≡b mod p−1 p -1 p−1. The point is that the list of possible orders of elements of Z2p+1∗ {\mathbb Z}_{2p+1}^* Z2p+1∗ is very short: the order of 2 2 2 divides ϕ(2p+1)=2p \phi(2p+1) = 2p ϕ(2p+1)=2p, so it is either 1,2,p, 1,2,p,1,2,p, or 2p2p2p. That is, every element b∈Zn∗ b \in {\mathbb Z}_n^* b∈Zn∗ can be written as gx g^x gx mod n n n, for some integer x x x. Here is a table of their powers modulo 14: A generator of (Z=p) is called a primitive root mod p. Example: Take p= 7. Zn∗={a∈N :1≤a Samsung Dryer Keeps Burning Heating Element,
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