anonymous
  • anonymous
Find all primitive roots of 41 and 82.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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misssunshinexxoxo
  • misssunshinexxoxo
This should help https://answers.yahoo.com/question/index?qid=20150407193751AAnlAo6
anonymous
  • anonymous
If you knew that 41 is a prime number, has no integer factors and that this number will divide into 82, would that help?
anonymous
  • anonymous
Is there a formula that would help me find the roots? @robtobey

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More answers

ganeshie8
  • ganeshie8
It is easy to find all the primitive roots after finding one primitive root
ganeshie8
  • ganeshie8
If \(n\) 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
  • ganeshie8
So you may try finding one primitive root by trial and error
anonymous
  • anonymous
@ganeshie8 Would you please show me an example.
ganeshie8
  • ganeshie8
Lets work primitive roots of 41
ganeshie8
  • ganeshie8
\(\phi(41) = 40\), so if \(a\) is a primitive root, then the order of \(a\) must be \(40\)
anonymous
  • anonymous
So there are 16 primitive roots
UsukiDoll
  • UsukiDoll
can we not post links to yahoo answers? What is the point? thanks.
ganeshie8
  • ganeshie8
There 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
  • ganeshie8
\(\large 3\) : \(3^8 \equiv 81^2 \equiv 1 \pmod{41}\) so this is a fail too
ganeshie8
  • ganeshie8
\(\large 5\) : \(5^{20} \equiv (5^5)^4 \equiv 9^4 \equiv 81^2 \equiv 1 \pmod{41}\) so this is a fail too
ganeshie8
  • ganeshie8
\(\large 6\) : \(6^{20} \equiv (6^2)^{10} \equiv (-5)^{10} \equiv 9^2 \equiv \color{red}{-1} \pmod{41}\) 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
  • ganeshie8
see if everything makes sense so far, all we did is tested each interger successively for primitive root
ganeshie8
  • ganeshie8
And yes there are exactly \(\phi(\phi(41)) = 16\) incongruent primitive roots for \(41\)
anonymous
  • anonymous
Then how would we use 6 to find the rest @ganeshie8
ganeshie8
  • ganeshie8
Since \(6\) 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
  • anonymous
I think I can follow your steps to get for 82. Thanks @ganeshie8
ganeshie8
  • ganeshie8
Yw, 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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