How many incongruent primitive roots does 13 have?
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How many incongruent primitive roots does 13 have?:
So far I know that phi(phi(13)) = phi(12) = 4, so there would be 4 primitive roots. And a RRS mod 13 is {1,2,3,...,12}. But I'm not sure where to go from here to find the incongruent primitive roots. Any help would be appreciated!
number-theory
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
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add a comment |
$begingroup$
How many incongruent primitive roots does 13 have?:
So far I know that phi(phi(13)) = phi(12) = 4, so there would be 4 primitive roots. And a RRS mod 13 is {1,2,3,...,12}. But I'm not sure where to go from here to find the incongruent primitive roots. Any help would be appreciated!
number-theory
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
$endgroup$
2
$begingroup$
$13$ is small, you could brute-force it. But if you have found one primitive root, do you know how to get the others from it?
$endgroup$
– Daniel Fischer♦
Feb 4 '16 at 0:52
$begingroup$
The only candidates are the quadratic non-residues. That cuts down on the work. There are further shortcuts, which are not worth using here.
$endgroup$
– André Nicolas
Feb 4 '16 at 1:43
add a comment |
$begingroup$
How many incongruent primitive roots does 13 have?:
So far I know that phi(phi(13)) = phi(12) = 4, so there would be 4 primitive roots. And a RRS mod 13 is {1,2,3,...,12}. But I'm not sure where to go from here to find the incongruent primitive roots. Any help would be appreciated!
number-theory
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
$endgroup$
How many incongruent primitive roots does 13 have?:
So far I know that phi(phi(13)) = phi(12) = 4, so there would be 4 primitive roots. And a RRS mod 13 is {1,2,3,...,12}. But I'm not sure where to go from here to find the incongruent primitive roots. Any help would be appreciated!
number-theory
number-theory
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
asked Feb 4 '16 at 0:49
whatarethosewhatarethose
285
285
2
$begingroup$
$13$ is small, you could brute-force it. But if you have found one primitive root, do you know how to get the others from it?
$endgroup$
– Daniel Fischer♦
Feb 4 '16 at 0:52
$begingroup$
The only candidates are the quadratic non-residues. That cuts down on the work. There are further shortcuts, which are not worth using here.
$endgroup$
– André Nicolas
Feb 4 '16 at 1:43
add a comment |
2
$begingroup$
$13$ is small, you could brute-force it. But if you have found one primitive root, do you know how to get the others from it?
$endgroup$
– Daniel Fischer♦
Feb 4 '16 at 0:52
$begingroup$
The only candidates are the quadratic non-residues. That cuts down on the work. There are further shortcuts, which are not worth using here.
$endgroup$
– André Nicolas
Feb 4 '16 at 1:43
2
2
$begingroup$
$13$ is small, you could brute-force it. But if you have found one primitive root, do you know how to get the others from it?
$endgroup$
– Daniel Fischer♦
Feb 4 '16 at 0:52
$begingroup$
$13$ is small, you could brute-force it. But if you have found one primitive root, do you know how to get the others from it?
$endgroup$
– Daniel Fischer♦
Feb 4 '16 at 0:52
$begingroup$
The only candidates are the quadratic non-residues. That cuts down on the work. There are further shortcuts, which are not worth using here.
$endgroup$
– André Nicolas
Feb 4 '16 at 1:43
$begingroup$
The only candidates are the quadratic non-residues. That cuts down on the work. There are further shortcuts, which are not worth using here.
$endgroup$
– André Nicolas
Feb 4 '16 at 1:43
add a comment |
1 Answer
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The four incongruent primitive roots of $13$ are $g^k$ for $g$ a primitive root and $k$ coprime with $12$, that is, $k=1,5,7,11$. They can also be given by $pm g, pm g^5$ because $g^6=-1$.
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
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add a comment |
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$begingroup$
The four incongruent primitive roots of $13$ are $g^k$ for $g$ a primitive root and $k$ coprime with $12$, that is, $k=1,5,7,11$. They can also be given by $pm g, pm g^5$ because $g^6=-1$.
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
$endgroup$
add a comment |
$begingroup$
The four incongruent primitive roots of $13$ are $g^k$ for $g$ a primitive root and $k$ coprime with $12$, that is, $k=1,5,7,11$. They can also be given by $pm g, pm g^5$ because $g^6=-1$.
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
$endgroup$
add a comment |
$begingroup$
The four incongruent primitive roots of $13$ are $g^k$ for $g$ a primitive root and $k$ coprime with $12$, that is, $k=1,5,7,11$. They can also be given by $pm g, pm g^5$ because $g^6=-1$.
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
$endgroup$
The four incongruent primitive roots of $13$ are $g^k$ for $g$ a primitive root and $k$ coprime with $12$, that is, $k=1,5,7,11$. They can also be given by $pm g, pm g^5$ because $g^6=-1$.
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
edited Dec 26 '18 at 10:23
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
answered Feb 4 '16 at 1:12
lhflhf
166k10171400
166k10171400
add a comment |
add a comment |
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$begingroup$
$13$ is small, you could brute-force it. But if you have found one primitive root, do you know how to get the others from it?
$endgroup$
– Daniel Fischer♦
Feb 4 '16 at 0:52
$begingroup$
The only candidates are the quadratic non-residues. That cuts down on the work. There are further shortcuts, which are not worth using here.
$endgroup$
– André Nicolas
Feb 4 '16 at 1:43