You're asking why 2 isn't also a primitive root? Lv 4. The primitive roots are 3, 10, 5, 11, 14, 7, 12, and 6. Modulo 11. Primitive Roots Calculator. 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). 28 is congruent to 1 mod 17. 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. i know, the homework problem was to find all 8, but i was just wondering why 2 didn't work. φ(φ(17)) = φ(16) = 8, so there are 8 primitive roots. 216 = 12 + 17*12, so 216 is congruent to 12 mod 17. Primitive Root Video. thanks! Something like, oh I don't know, http://www.reddit.com/r/cheatatmathhomework. i'm working on some number theory homework and i didn't know who else to ask since it's late at night. thanks! By using our Services or clicking I agree, you agree to our use of cookies. You're asking why 2 isn't also a primitive root? Cookies help us deliver our Services. Here is a table of their powers modulo 14: Thus the powers of 2 from 1 to 16 won't form the desired complete reduced residue class. Menu. i know 316 = 1 mod 17, but isn't 216 = 1 mod 17 as well? 11 has phi(10) = 4 primitive roots. These are 8,7,and6 . that makes sense. Smallest primitive root is the smallest positive number that is a primitive root modulo a given number. numbers are prime to 10. Primitive Root Calculator-- Enter p (must be prime)-- Enter b . The primitive roots are 3, 11, 12, 13, 17 ≡ –14, 21 ≡ –10, 22 ≡ –9, and 24 ≡ –7. 4- If it is 1 then 'i' is not a primitive root of n. 5- If it is never 1 then return i;. Thus the powers of 2 from 1 to 16 won't form the desired complete reduced residue class. 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. 3 0. hisamuddin. the others are in positions whose position. 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. i didn't know there was one. 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. 4 years ago. 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 Calculator. Their product 970377408 ≡ 1 (mod 31) and their sum 123 ≡ –1 (mod 31). Primitive Root Calculator. Finding Other Primitive Roots (mod p) Suppose that we have a primitive root, g. For example, 2 is a primitive root of 59. definitely will use that in the future. 2: 2,4,8,5,10,9,7,3,6,1 so 2 is a primitive root. The first 10,000 primes, if you need some inspiration. 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. 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 primitive roots are 3, 11, 12, 13, 17 ≡ –14, 21 ≡ –10, 22 ≡ –9, and 24 ≡ –7. Alternate Solution : Observing φ(17) = 16, if a is reduced modulo 17 then ord17 a ∈ {1,2,4,8,16}. 0 0. leister. Correct me if I'm misinterpreting your question. 2 8 is congruent to 1 mod 17. 3×11 = 33 ≡ 2 It follows immediately that (1) is a complete listing of the primitive roots of 17. http://www.reddit.com/r/cheatatmathhomework. 2^16 = 65536 which is congruent to 1 mod(17) Which means it should be a primitive root of 17. Press question mark to learn the rest of the keyboard shortcuts. Other related properties. 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. Lv 4. Given that 2 is a primitive root of 59, find 17 other primitive roots … so the primitive roots are 2,6,7,8. You must have made a mistake in your arithmetic. incongruent primitive roots of 17. Source(s): https://shorte.im/bagFW. 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. 4 years ago. Email: donsevcik@gmail.com Tel: 800-234-2933; Their product 970377408 ≡ 1 (mod 31) and their sum 123 ≡ –1 (mod 31). Answer #2 | 19/04 2015 19:12 In answering an earlier question, I showed that 3 is a primitive root of 17. Example 1. Finding primitive roots . If only we and a subreddit for that. (you can find all of them by taking odd powers of 3, if you want). 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. Let's test. Enter a prime number into the box, then click "submit." am i missing something? Smallest magnitude primitive root is the primitive root with the smallest absolute value. It will calculate the primitive roots of your number.