primitive roots of 17

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November 29th, 2020

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

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