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anonymous
 one year ago
Find all primitive roots of 41 and 82.
anonymous
 one year ago
Find all primitive roots of 41 and 82.

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misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.1This should help https://answers.yahoo.com/question/index?qid=20150407193751AAnlAo6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you knew that 41 is a prime number, has no integer factors and that this number will divide into 82, would that help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there a formula that would help me find the roots? @robtobey

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It is easy to find all the primitive roots after finding one primitive root

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3If \(n\) has a primitive root, then it has exactly \(\phi(\phi(n))\) primitive roots. These can be found trivially by using any known primitive root.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3So you may try finding one primitive root by trial and error

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Would you please show me an example.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Lets work primitive roots of 41

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\phi(41) = 40\), so if \(a\) is a primitive root, then the order of \(a\) must be \(40\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So there are 16 primitive roots

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0can we not post links to yahoo answers? What is the point? thanks.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3There is no known analytic method for finding a primitive root. Once you find one primitive root, there is an easy way to find all other primitive roots. So lets go ahead and find one primitive root by work each integer : \(\large 2\) : \(2^{20} \equiv (2^5)^4\equiv (9)^4 \equiv 81^2\equiv (1)^2\equiv 1 \pmod{41}\) That means the order of \(2\) must be \(\le 20\). so \(2\) cannot be a primitive root of \(41\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\large 3\) : \(3^8 \equiv 81^2 \equiv 1 \pmod{41}\) so this is a fail too

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\large 5\) : \(5^{20} \equiv (5^5)^4 \equiv 9^4 \equiv 81^2 \equiv 1 \pmod{41}\) so this is a fail too

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\large 6\) : \(6^{20} \equiv (6^2)^{10} \equiv (5)^{10} \equiv 9^2 \equiv \color{red}{1} \pmod{41}\) Also \(6^{8} \equiv (6^2)^4 \equiv (5)^4 \equiv \color{red}{10} \pmod{41}\) so we got lucky just now, \(6\) is a primitive root of \(41\). we can go use this to find all other primitive roots

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3see if everything makes sense so far, all we did is tested each interger successively for primitive root

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3And yes there are exactly \(\phi(\phi(41)) = 16\) incongruent primitive roots for \(41\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then how would we use 6 to find the rest @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Since \(6\) is a primitive root of \(41\), any coprime intger is congruent to an integer of form \(6^{k}\), where \(1\le k\le 30\). You must be knowing that the order of \(6^k\) is given by \[\dfrac{41}{\gcd(k,~40)}\] This will equal \(41\) if and only if \(\gcd(k, 40)=1\). so \(k \in\{1,3,7,9,11,13,17,19,21,23,27,29,31,33,37,39\}\) and the corresponding primitive roots are \(\{6^{1},6^{3},6^{7},6^{9},6^{11},6^{13},6^{17},6^{19},6^{21},6^{23},6^{27},6^{29},6^{31},6^{33},6^{37},6^{39}\}\) you may reduce them under mod 41 if you wish

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I can follow your steps to get for 82. Thanks @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yw, as you can see almost all the effort was spent finding one primitive root, once you know one primitive root, generating the remaining is trivial.
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