Randomly? Copyright © 2005-2020 Math Help Forum. Mathematics is concerned with numbers, data, quantity, structure, space, models, and change. $1$ cannot be a primitive root. Also, there will be more powers to test. Once you have found one primitive root, you can easily find all the others. If $2^2$ was one, then 2 would've been one as well. You could pick the candidates randomly. No simple general formula to compute primitive roots … So the only number we don't need to check is 5 as it is not coprime to 25. As it happens, sums (or differences) of two primitive roots add up to all elements of the index 2 subgroup of Z/n Z for even n, and to the whole group Z/n Z when n is odd: Z/n Z × + Z/n Z × = Z/n Z or 2Z/n Z. Let $p$ be a prime with $p\equiv 1\pmod4$, and let $g$ be a primitive root modulo $p.$ Then $-1\equiv g^{\frac{p-1}{2}}\pmod p,$ so $-g\equiv g^{\frac{p+1}{2}}\pmod p.$ Since $\frac{p+1}{2}$ is prime to $p-1,$ $-g$ is also a primitive root modulo $p.$, Finding a primitive root of a prime number, “Question closed” notifications experiment results and graduation, MAINTENANCE WARNING: Possible downtime early morning Dec 2/4/9 UTC (8:30PM…, Generator of multiplicative group of $\mathbb{Z}/p\mathbb{Z}$, Efficient algorithms for Primitive roots where time-complexity is $\leq O(\sqrt{n})$. Finally, calculate $a^{s/p_i}\mod p$ for all $i=1\ldots k$, and if you find $1$ among residuals then it is NOT a primitive root, otherwise it is. I think composite numbers are a bit less likely to be primitive roots when their factors aren't (someone will know the details). Right. 1 Answer. 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.. How would you find a primitive root of a prime number such as 761? See what happens to $3$ when you test it against $761$: $3^{380}\equiv -1\mod 761$, but, in fact, $3^{380/5}=3^{76}\equiv -1\mod 761$ as well, and $3^{152}=1\mod 761$. For example, if you pick a random number to test for being a primitive root of $761$, then the probability of finding one is roughly $\frac{1}{2}\times\frac{4}{5}\times\frac{18}{19}$ or 38%, and there are 3 powers to test. What is the minimum viable ecological pyramid a terrafoming project would introduce to world with no life to make it suitable for humans? Find all primitive roots of $71$ and also find a primitive root for $p^2$ and for $2p^2$. Does it have to do with the fact that it's $2^2$? All rights reserved. That removes $1, 4, 8, 9$ and others. Typically, what you do is you pick a number and test. Finding primitive roots. For example, $6^2=36$ or $6^{15}\equiv 686$ are not primitive roots of $761$ because $\gcd(2,760)=2>1$ and $\gcd(15,760)=5>1$, but, for example, $6^3=216$ is another primitive root of 761. Once you have found one primitive root, you can easily find all the others. if i don't misunderstand ,the way you say is if i replace 2 by another number i and \(\displaystyle (p-1)/i\) is prime ,then i is primitive root of n . 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. 1 decade ago. 25 = 5^5. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. JavaScript is disabled. But if you are looking for primitive roots of, say, $2311$ then the probability of finding one at random is about 20% and there are 5 powers to test. Do all threads share the same instance of a heap variable, or have different instances of a heap variable? A quadratic residue cannot be a primitive root of an odd prime. maybe i am not smart enough to know your advice.YOu can tell clearly . How can I find the area of an overlayer structure? Relevance. This can be proved as follows. There are primitive roots mod n n n if and only if n = 1, 2, 4, p k, n = 1,2,4,p^k, n = 1, 2, 4, p k, or 2 p k, 2p^k, 2 p k, where p p p is an odd prime. And $a^m \mod p$ is another primitive root if and only if $m$ and $p-1$ are coprime (if $\gcd(m,p-1)=d$ then $(a^m)^{(p-1)/d}\equiv (a^{p-1})^{m/d}\equiv 1\mod p$, so we need $d=1$). Why does Chrome need access to Bluetooth? For $761$, there are exactly $\phi(\phi(761)) = \phi(760) = \phi(2^3\times 5\times 19) = 2^2\times 4\times 18 = 288$ primitive roots, so you have about a 3/8 change of picking a primitive root by picking one at random. For a better experience, please enable JavaScript in your browser before proceeding. To test that $a$ is a primitive root of $p$ you need to do the following. Favorite Answer. Meaning of the Term "Heavy Metals" in CofA? what are the eight primitive roots of 25, how can you tell? If $p-1$ is divisible by $4$ and $a$ is a primitive element then $p-a$ is also primitive. In fact, when you test $3$ only power $152$ (among the three powers above) will tell you that it is not a primitive root, when you test 35 -- only 40, and when test 2 -- only 380. Why do internal forces not affect the conservation of momentum, the powers to test are: $760/2=380$, $760/5=152$ and $760/19=40$ (just 3 instead of testing all of them). By using primitive roots how does one solve $x^2 \equiv -1 \pmod p$ for $x$, given prime $p$. Anonymous. Is There (or Can There Be) a General Algorithm to Solve Rubik's Cubes of Any Dimension? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. I think if testing all i ,(i,N)=1, is so long THanks The prime $ p = 71$ has $7$ as a primitive root. Given that 2 is a primitive root of 59, find 17 other primitive roots of 59. 4- If it is 1 then 'i' is not a primitive root of n. 5- If it is never 1 then return i;. How you find all the other primitive roots. Why did mainframes have big conspicuous power-off buttons? Let N have primitive roots .Which the way does find all primitive roots generally ? Now since you one primitive root you found them all. If a person is dressed up as non-human, and is killed by someone who sincerely believes the victim was not human, who is responsible? Then you need to determine all the prime factors of $s$: $p_1,\ldots,p_k$. You do not raise $a$ to the power of all prime factors of $\phi(p)$. Finding primitive roots is generally difficult. Has anyone seriously considered a space-based time capsule? You find those factors $p_i$ and raise $a$ to the powers $s/p_i$. A given $a$ is a primitive root modulo $761$ if and only if the order of $a$ modulo $761$ is, @Arturo This is a necessary condition, but not sufficient. I think if testing all i ,(i,N)=1, is so long. Since you can store quite large powers of small primes in a table, calculating their powers will be a little bit faster. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. You're not just testing primes, though, as you find one at 6.. @Joost Yes, being a perfect power prevents an integer from being the smallest(!) Finding Primitive Roots The proof of the theorem (part of which is presented below) is essentially non-constructive: that is, it does not give an effective way to find a primitive root when it exists. Of course, a computer program is the best way to go. Once you find one primitive root, you find all the others. So I'd check small primes first. How to find individual probabilities of all numbers from a list? Let us find the lowest primitive root of $761$: So, the least primitive root of 761 is 6. Say \(\displaystyle p=11\) then \(\displaystyle (11-1)/2=5\) is prime, which means \(\displaystyle 2\) is primitive root of \(\displaystyle 11\). For humans has $ 7 $ as a primitive root of an overlayer structure,. Primes in a table, calculating their powers will be a primitive root, can... The wing of Embraer ERJ-145 've been one as well powers to test all the prime $ p 3! In related fields paste this URL into your RSS reader creature in existence $ s=p-1 $ to know advice.YOu! 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