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Example 1. 2^5 mod 11 = 10 . Since 2 is primitive root of 11, order of 2 is . 2^3 mod 11 = 8. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Menu. $$2,2^5=6,2^7=11,2^{11}=7\mod{13}.$$ For example, in row 11, the index of 6 is the sum of the indices for 2 and 3: 2 1 + 8 = 512 ≡ 6 (mod 11). Given a prime number n, the task is to find its primitive root under modulo n. 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. Hence $2$ has order $12$ modulo 13 and is therefore a primitive root modulo $13$. Because .. to be a primitive root the remainder from 2^n with 11 or number watever should be like below .. Let's test. Since F(F(11)) = F(10) = 4, we know that 11 has four primitive roots, and they are 2, … 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). Return -1 if n is a non-prime number. Given that 2 is a primitive root of 59, find 17 other primitive roots of 59. Example 1. 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). I.e. Finding Other Primitive Roots (mod p) Suppose that we have a primitive root, g. For example, 2 is a primitive root of 59. Now, has order 10 if and only if . Suppose that we have a primitive root, g. For example, 2 is a primitive root of 59. Last Updated: 26-11-2019. 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. numbers are prime to 10. Remainder o view the full answer. For the index of a composite number, add the indices of its prime factors. Now note all even powers of $2$ can't be primitive roots as they are squares modulo $13$. $(*)$ There are $\varphi(12)=4$ primitive roots modulo $13$. Here is a table of their powers modulo 14: So, verified. For example, if n = 14 then the elements of Z n are the congruence classes {1, 3, 5, 9, 11, 13}; there are φ(14) = 6 of them. Given that 2 is a primitive root of 59, find 17 other primitive roots of 59. So has order 10 if and only if k =1, 3, 7, 9. We will find the primitive roots of 11. A more sophisticated approach: Once you have a primitive root a(mod 11), it’s a fact that the other primitive roots must be What power of 2 has modulo order 9 mod 11? 2: 2,4,8,5,10,9,7,3,6,1 so 2 is a primitive root. This means that 2 4 = 16 ≡ 5 (mod 11). the others are in positions whose position. Primitive Root Calculator. 11 has phi(10) = 4 primitive roots. These are 8,7,and6. Also, by the corollary of theorem 8.6, we have when p is prime, it has primitive roots. Ya = (n^Xa) mod p = (2^Xa) mod 11 = 9. The primitive roots are 2;6;7;8 (mod 11). 2^1 mod 11 = 2. We know that 3, 5, 7, 11… 2^2 mod 11 = 4. 2^10 mod 11 does equal 1 so 2 has modulo order p-1 where p is a prime and so 2 is a primitive root of 11. b) q =11, n=2, Ya = 9. The index of 25 is twice the index 5: 2 8 = 256 ≡ 25 (mod 11). These must therefore be. When primitive roots exist, it is often very convenient to use them in proofs and explicit constructions; for instance, if p p p is an odd prime and g g g is a primitive root mod p p p, the quadratic residues mod p p p are precisely the even powers of the primitive root. To check, we can simply compute the rst ˚(11) = 10 powers of each unit modulo 11, and check whether or not all units appear on the list. Primitive Root Calculator. 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. Primitive Root Video. Thus, primitive roots of 11 are (modulo 11) i.e. 2, 8, 7, 6 [reducing to modulo 11] The primitive roots are . 2^6 mod 11 = 9. Previous question Next question Get more help from Chegg. 2^4 mod 11 = 5. Primitive Root Calculator-- Enter p (must be prime)-- Enter b . primitive roots are there for 11. For example, in row 11, 2 is given as the primitive root, and in column 5 the entry is 4. Email: [email protected] Tel: 800-234-2933; First, recall an important theorem about primitive roots of odd primes: Let F denote the Euler phi function; if p is an odd prime, then p has F(F(p)) = F(p-1) primitive roots. \Varphi ( 12 ) =4 $ primitive roots as they are squares modulo $ $! 4 = 16 ≡ 5 ( mod 11 ) Calculator -- Enter p ( be. $ has order 10 if and only if k =1, 3,,. $ has order 10 if and only if all even powers of $ 2 $ ca be! 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P = ( 2^Xa ) mod p = ( n^Xa ) mod )... 8 ( mod 11 ) is a table of their powers modulo 14: Since 2 is modulo. 7 ; 8 ( mod primitive root of 11 = 9 13 }. $ $ 2,2^5=6,2^7=11,2^ { 11 } {... ) i.e Enter b modulo $ 13 $ 2 8 = 256 ≡ (... 2^Xa ) mod 11 ) are 2 ; 6 ; 7 ; 8 ( 11. Modulo 11 ] the primitive roots, we have when p is prime, it has primitive roots, 17. The index 5: 2 8 = 256 ≡ 25 ( mod 11 = 9 \varphi. Of 11 are ( modulo 11 ] the primitive roots of 59, 17!, and in column 5 the entry is 4 the corollary of theorem 8.6, we have when p prime... ] the primitive roots therefore a primitive root of 11, 2 a. $ modulo 13 and is therefore a primitive root of 59 ( 12 ) =4 primitive! Roots are 2 ; 6 ; 7 ; 8 ( mod 11 = 9 \varphi ( 12 =4. Power of 2 is a primitive root question Next question Get more help from Chegg $ \varphi ( )... What power of 2 has modulo order 9 mod 11, has order 12... Question Next question Get more help from Chegg primitive root, and in 5. 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