The standard theorem here is that if a is a primitive root of p 2, where p is prime, then a is a primitive root of p k for any k ≥ 2. In the following theorem, we prove that no power of 2, other than 2 or 4, has a primitive root and that is because when \(m\) is an odd integer, \(ord_2^km\neq \phi(2^k)\) and this is because \(2^k\mid (a^{\phi(2^k)/2}-1)\). I see. Prove that 3 is a primitive root of $7^k$ for all $k \ge 1$, math.stackexchange.com/questions/594782/…, math.stackexchange.com/questions/332760/…, “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…. Let’s say, means is a primitive root . In fact, ɸ (7) = 6, since the numbers 3 1 – 1 = 2, 3 2 – 1 = 8, 3 3 - 1 = 26, 3 4 - 1 = 80, and 3 5 - 1 = 242 are not divisible by 7—only 3 6 — 1 = 728 is divisible by 7. Definition 5 (Legendre Symbol): is called the Legendre symbol for a prime . Definition 4 (Quadratic Residue): is a of if. The polynomial ∏ ζ a primitive n th root of unity (x − ζ) \prod_{\zeta \text{ a primitive } n\text{th root of unity}} (x-\zeta) ζ a primitive n th root of unity ∏ (x − ζ) is a polynomial in x x x known as the n n n th cyclotomic polynomial. To learn more, see our tips on writing great answers. Question: Is 3 A Primitive Root Of 7? The standard theorem here is that if $a$ is a primitive root of $p^2$, where $p$ is prime, then $a$ is a primitive root of $p^k$ for any $k\ge 2$. All content on this website, including dictionary, thesaurus, literature, geography, and other reference data is for informational purposes only. ]](x) denote the number of square-free, (6) a.sup.m-1 mod m = 1 (7) and a standard number theoretic definition: If m is prime then a is a primitive element modulo m (or, By Horie [4, Theorem 2], l [??] Is There (or Can There Be) a General Algorithm to Solve Rubik's Cubes of Any Dimension? Prove that number is a primitive root for all $k$ in range. The possible orders are multiples of $6$ that divide $42$. Answered October 31, 2017. so we need only show that $3$ does not have order $6$ modulo $7^2$. A primitive root mod n n n is an integer g g g such that every integer relatively prime to n n n is congruent to a power of g g g mod n n n. That is, the integer g g g is a primitive root (mod n n n) if for every number a a a relatively prime to n n n there is an integer z z z such that a ≡ (g z (m o d n)). Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. I am trying to prove this via induction. Return -1 if n is a non-prime number. But the only generators modulo $p^{k+1}$ are generators modulo $p^k$, and there are only $pa_k$ numbers, modulo $p^{k+1}$ which are congruent to generators modulo $p^{k}$. Justify Your Answer ( Show Step By Step Working Out) This problem has been solved! In fact, ɸ(7) = 6, since the numbers 31 – 1 = 2, 32 – 1 = 8, 33 - 1 = 26, 34 - 1 = 80, and 35 - 1 = 242 are not divisible by 7—only 36 — 1 = 728 is divisible by 7. Menu. The first few for which primitive roots exist are 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 17, 18, 19, 22, ... (OEIS A033948 ), so the number of primitive root of order for , 2, ... are 0, 1, 1, 1, 2, 1, 2, 0, 2, 2, 4, 0, 4, ... (OEIS A046144 ). Looking for a function that approximates a parabola. For example, if m = 7, the number 3 is a primitive root modulo 7. In 1926, 1. The number 3 is a primitive root modulo 7 [1] because. It is of great interest in algebraic number theory. Why would I choose a bike trainer over a stationary bike? Thanks for contributing an answer to Mathematics Stack Exchange! rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. First, use that the multiplicative group $\mathbb Z_{p^k}^\times$ is cyclic. 3) For each primitive root b in the table, b 0, b 1, b 2 is a primitive root mod $3^h$ for any positive integer $h$, if $b^k$ is a primitive root, then $b$ is a primitive root. Explanation: n = 7. Use MathJax to format equations. Why did mainframes have big conspicuous power-off buttons? Smallest primitive root = 3. This is in principle a computation, but we can speed it up. 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