Modulo 11. Thus the powers of 2 from 1 to 16 won't form the desired complete reduced residue class. thanks! Thus the powers of 2 from 1 to 16 won't form the desired complete reduced residue class. Answer #2 | 19/04 2015 19:12 In answering an earlier question, I showed that 3 is a primitive root of 17. Primitive Root Calculator. Primitive Roots Calculator. 11 has phi(10) = 4 primitive roots. 0 0. leister. Cookies help us deliver our Services. These are 8,7,and6 . the others are in positions whose position. 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. Although there can be multiple primitive root for a prime number but we are only concerned for smallest one.If you want to find all roots then continue the process till p-1 instead of breaking up on finding first primitive root. Finding primitive roots . The primitive roots are 3, 11, 12, 13, 17 ≡ –14, 21 ≡ –10, 22 ≡ –9, and 24 ≡ –7. Lv 4. I have plugged through the definition of the primitive root of 17, Phi(17) = 16. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. The first 10,000 primes, if you need some inspiration. Enter a prime number into the box, then click "submit." definitely will use that in the future. so the primitive roots are 2,6,7,8. 3×11 = 33 ≡ 2 thanks! Menu. By using our Services or clicking I agree, you agree to our use of cookies. incongruent primitive roots of 17. 2 8 is congruent to 1 mod 17. 2^16 = 65536 which is congruent to 1 mod(17) Which means it should be a primitive root of 17. (you can find all of them by taking odd powers of 3, if you want). Smallest primitive root is the smallest positive number that is a primitive root modulo a given number. Finding Other Primitive Roots (mod p) Suppose that we have a primitive root, g. For example, 2 is a primitive root of 59. Something like, oh I don't know, http://www.reddit.com/r/cheatatmathhomework. Lv 4. i know, the homework problem was to find all 8, but i was just wondering why 2 didn't work. You must have made a mistake in your arithmetic. 2: 2,4,8,5,10,9,7,3,6,1 so 2 is a primitive root. Press question mark to learn the rest of the keyboard shortcuts. Primitive Root Video. 4 years ago. Email: donsevcik@gmail.com Tel: 800-234-2933; 4 years ago. You're asking why 2 isn't also a primitive root? 3 0. hisamuddin. If only we and a subreddit for that. Their product 970377408 ≡ 1 (mod 31) and their sum 123 ≡ –1 (mod 31). For it to be a primitive root of p, it's required that the smallest value of h such that 2 h is congruent to 1 mod p be p - 1. crumunch beat me to the punch, but I'll add that saying "the" primitive root is very much the wrong way to think about it. Primitive Root Calculator. 28 is congruent to 1 mod 17. The primitive roots are 3, 11, 12, 13, 17 ≡ –14, 21 ≡ –10, 22 ≡ –9, and 24 ≡ –7. Other related properties. i didn't know there was one. Their product 970377408 ≡ 1 (mod 31) and their sum 123 ≡ –1 (mod 31). i'm working on some number theory homework and i didn't know who else to ask since it's late at night. Primitive Root Calculator-- Enter p (must be prime)-- Enter b . Given that 2 is a primitive root of 59, find 17 other primitive roots … i guess it copied incorrectly, it was supposed to say 316 and 216, i'll go fix that, New comments cannot be posted and votes cannot be cast, Press J to jump to the feed. that makes sense. Here is a table of their powers modulo 14: 4- If it is 1 then 'i' is not a primitive root of n. 5- If it is never 1 then return i;. 216 = 12 + 17*12, so 216 is congruent to 12 mod 17. http://www.reddit.com/r/cheatatmathhomework. Correct me if I'm misinterpreting your question. 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. Alternate Solution : Observing φ(17) = 16, if a is reduced modulo 17 then ord17 a ∈ {1,2,4,8,16}. 3×11 = 33 ≡ 2 12×13 = 156 ≡ 1 (–14)×(–10) = 140 ≡ 16 (–9)×(–7) = 63 ≡ 1, and 2×1×16×1 = 32 ≡ 1 (mod 31). For it to be a primitive root of p, it's required that the smallest value of h such that 2h is congruent to 1 mod p be p - 1. oh, it's because 28 hits 1 before 216 can. The number of primitive roots modulo , if the multiplicative group is cyclic, ... 17 : 8 : 3,-3 : 3 : 3,5,6,7,10,11,12,14 Relation with other properties Smallests. numbers are prime to 10. φ(φ(17)) = φ(16) = 8, so there are 8 primitive roots. am i missing something? Let's test. Smallest magnitude primitive root is the primitive root with the smallest absolute value. i know 316 = 1 mod 17, but isn't 216 = 1 mod 17 as well? It follows immediately that (1) is a complete listing of the primitive roots of 17. The primitive roots are 3, 10, 5, 11, 14, 7, 12, and 6. When primitive roots exist mod n, (see my blog post here for a breakdown of when they do or do not exist) there will be exactly φ(φ(n)) of them. Source(s): https://shorte.im/bagFW. It will calculate the primitive roots of your number. You're asking why 2 isn't also a primitive root? Example 1.