Given 2 is a primitive root mod 19, find all solutions to x^12 ≡ 7 (mod 19) (a) x^12 ≡ 6 (mod 19) The 2019 Stack Overflow Developer Survey Results Are InFinding all solutions to $3^x equiv 9 pmod13$$r$ primitive root of prime $p$, where $p equiv 1 mod 4$: prove $-r$ is also a primitive rootAt least $p^2-p$ solutions to $x^2+y^2+z^2 equiv 1 mod p$if $b^k$ is a primitive root, then $b$ is a primitive rootDetermine the number of solutions of $x^pequiv 1mod p^h$ using primitive rootsShow that 3 is a primitive root modulo 14. Then, write the other primitive roots modulo 14 in terms of powers of 3. How many are there?primitive root modulo prime powersShow that there is a primitive root mod mEvaluate sum over primitive root mod pShowing two different definitions of a primitive root are the same
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Given 2 is a primitive root mod 19, find all solutions to x^12 ≡ 7 (mod 19) (a) x^12 ≡ 6 (mod 19)
The 2019 Stack Overflow Developer Survey Results Are InFinding all solutions to $3^x equiv 9 pmod13$$r$ primitive root of prime $p$, where $p equiv 1 mod 4$: prove $-r$ is also a primitive rootAt least $p^2-p$ solutions to $x^2+y^2+z^2 equiv 1 mod p$if $b^k$ is a primitive root, then $b$ is a primitive rootDetermine the number of solutions of $x^pequiv 1mod p^h$ using primitive rootsShow that 3 is a primitive root modulo 14. Then, write the other primitive roots modulo 14 in terms of powers of 3. How many are there?primitive root modulo prime powersShow that there is a primitive root mod mEvaluate sum over primitive root mod pShowing two different definitions of a primitive root are the same
$begingroup$
I'm stuck on (another) problem under Number Theory. There are quite a few gaps on what was covered in my class so I'm having quite a bit of trouble. Could you please help me solve the following problem?
2 is a primitive root mod 19. Using this information, find all solutions to x^12 ≡ 7 (mod 19) and x^12 ≡ 6 (mod 19)
I think I would have to make use of the powers of 2 to solve this, but I can't get any further than that.
Any help would be much appreciated!
number-theory elementary-number-theory primitive-roots
asked Mar 23 at 19:07
KamiKami
13
$endgroup$
add a comment |
$begingroup$
I'm stuck on (another) problem under Number Theory. There are quite a few gaps on what was covered in my class so I'm having quite a bit of trouble. Could you please help me solve the following problem?
2 is a primitive root mod 19. Using this information, find all solutions to x^12 ≡ 7 (mod 19) and x^12 ≡ 6 (mod 19)
I think I would have to make use of the powers of 2 to solve this, but I can't get any further than that.
Any help would be much appreciated!
number-theory elementary-number-theory primitive-roots
asked Mar 23 at 19:07
KamiKami
13
$endgroup$
$begingroup$
You could start by listing out the powers of 2 $pmod19$...
$endgroup$
– Dzoooks
Mar 23 at 19:28
add a comment |
$begingroup$
I'm stuck on (another) problem under Number Theory. There are quite a few gaps on what was covered in my class so I'm having quite a bit of trouble. Could you please help me solve the following problem?
2 is a primitive root mod 19. Using this information, find all solutions to x^12 ≡ 7 (mod 19) and x^12 ≡ 6 (mod 19)
I think I would have to make use of the powers of 2 to solve this, but I can't get any further than that.
Any help would be much appreciated!
number-theory elementary-number-theory primitive-roots
asked Mar 23 at 19:07
KamiKami
13
$endgroup$
I'm stuck on (another) problem under Number Theory. There are quite a few gaps on what was covered in my class so I'm having quite a bit of trouble. Could you please help me solve the following problem?
2 is a primitive root mod 19. Using this information, find all solutions to x^12 ≡ 7 (mod 19) and x^12 ≡ 6 (mod 19)
I think I would have to make use of the powers of 2 to solve this, but I can't get any further than that.
Any help would be much appreciated!
number-theory elementary-number-theory primitive-roots
number-theory elementary-number-theory primitive-roots
asked Mar 23 at 19:07
KamiKami
13
asked Mar 23 at 19:07
KamiKami
13
asked Mar 23 at 19:07
KamiKami
13
asked Mar 23 at 19:07
KamiKami
13
asked Mar 23 at 19:07
KamiKami
13
13
$begingroup$
You could start by listing out the powers of 2 $pmod19$...
$endgroup$
– Dzoooks
Mar 23 at 19:28
add a comment |
$begingroup$
You could start by listing out the powers of 2 $pmod19$...
$endgroup$
– Dzoooks
Mar 23 at 19:28
$begingroup$
You could start by listing out the powers of 2 $pmod19$...
$endgroup$
– Dzoooks
Mar 23 at 19:28
$begingroup$
You could start by listing out the powers of 2 $pmod19$...
$endgroup$
– Dzoooks
Mar 23 at 19:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
Write $x=2^p$ (why ?) then you get $2^12pequiv 2^ypmod19$ where $y$ correspond to the representation for $7$ and $6$.
Then you get solve $12pequiv ypmod18$ (why ?)
answered Mar 23 at 19:52
zwimzwim
12.7k832
$endgroup$
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input? –
$endgroup$
– Kami
Mar 24 at 7:58
$begingroup$
Yes, this is it, for $x^12=7$ you have $6$ solutions and for $x^12=6$ there are none.
$endgroup$
– zwim
Mar 24 at 10:32
add a comment |
$begingroup$
Hint:
So you're given that $2$ has order $18bmod 19$. In particular, $7equiv 2^6mod 19$.
Any solution can be written as $2^k$, where $k$ is unique module $18$. The equation can be rewritten as
$$2^12kequiv 2^6 mod 19iff 12kequiv 6mod 18iff 2kequiv 1mod 3.$$
Can you continue?
answered Mar 23 at 19:56
BernardBernard
124k741117
$endgroup$
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input?
$endgroup$
– Kami
Mar 24 at 5:30
$begingroup$
@Kami: For the first equation, you find $kequiv 2mod 3$, i.e. $k=2, 5, 8$, &c. As to the second equation, $12kequiv 14pmod18iff 6kequiv 7pmod 9$, and there's not necessarily a solution, as $6$ is not a unit mod. $9$. Indeed $6$ has only three multiples mod. $9$: $;6, 3$ and $0$.
$endgroup$
– Bernard
Mar 24 at 11:11
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Write $x=2^p$ (why ?) then you get $2^12pequiv 2^ypmod19$ where $y$ correspond to the representation for $7$ and $6$.
Then you get solve $12pequiv ypmod18$ (why ?)
answered Mar 23 at 19:52
zwimzwim
12.7k832
$endgroup$
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input? –
$endgroup$
– Kami
Mar 24 at 7:58
$begingroup$
Yes, this is it, for $x^12=7$ you have $6$ solutions and for $x^12=6$ there are none.
$endgroup$
– zwim
Mar 24 at 10:32
add a comment |
$begingroup$
Hint:
Write $x=2^p$ (why ?) then you get $2^12pequiv 2^ypmod19$ where $y$ correspond to the representation for $7$ and $6$.
Then you get solve $12pequiv ypmod18$ (why ?)
answered Mar 23 at 19:52
zwimzwim
12.7k832
$endgroup$
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input? –
$endgroup$
– Kami
Mar 24 at 7:58
$begingroup$
Yes, this is it, for $x^12=7$ you have $6$ solutions and for $x^12=6$ there are none.
$endgroup$
– zwim
Mar 24 at 10:32
add a comment |
$begingroup$
Hint:
Write $x=2^p$ (why ?) then you get $2^12pequiv 2^ypmod19$ where $y$ correspond to the representation for $7$ and $6$.
Then you get solve $12pequiv ypmod18$ (why ?)
answered Mar 23 at 19:52
zwimzwim
12.7k832
$endgroup$
Hint:
Write $x=2^p$ (why ?) then you get $2^12pequiv 2^ypmod19$ where $y$ correspond to the representation for $7$ and $6$.
Then you get solve $12pequiv ypmod18$ (why ?)
answered Mar 23 at 19:52
zwimzwim
12.7k832
answered Mar 23 at 19:52
zwimzwim
12.7k832
answered Mar 23 at 19:52
zwimzwim
12.7k832
answered Mar 23 at 19:52
zwimzwim
12.7k832
12.7k832
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input? –
$endgroup$
– Kami
Mar 24 at 7:58
$begingroup$
Yes, this is it, for $x^12=7$ you have $6$ solutions and for $x^12=6$ there are none.
$endgroup$
– zwim
Mar 24 at 10:32
add a comment |
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input? –
$endgroup$
– Kami
Mar 24 at 7:58
$begingroup$
Yes, this is it, for $x^12=7$ you have $6$ solutions and for $x^12=6$ there are none.
$endgroup$
– zwim
Mar 24 at 10:32
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input? –
$endgroup$
– Kami
Mar 24 at 7:58
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input? –
$endgroup$
– Kami
Mar 24 at 7:58
$begingroup$
Yes, this is it, for $x^12=7$ you have $6$ solutions and for $x^12=6$ there are none.
$endgroup$
– zwim
Mar 24 at 10:32
$begingroup$
Yes, this is it, for $x^12=7$ you have $6$ solutions and for $x^12=6$ there are none.
$endgroup$
– zwim
Mar 24 at 10:32
add a comment |
$begingroup$
Hint:
So you're given that $2$ has order $18bmod 19$. In particular, $7equiv 2^6mod 19$.
Any solution can be written as $2^k$, where $k$ is unique module $18$. The equation can be rewritten as
$$2^12kequiv 2^6 mod 19iff 12kequiv 6mod 18iff 2kequiv 1mod 3.$$
Can you continue?
answered Mar 23 at 19:56
BernardBernard
124k741117
$endgroup$
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input?
$endgroup$
– Kami
Mar 24 at 5:30
$begingroup$
@Kami: For the first equation, you find $kequiv 2mod 3$, i.e. $k=2, 5, 8$, &c. As to the second equation, $12kequiv 14pmod18iff 6kequiv 7pmod 9$, and there's not necessarily a solution, as $6$ is not a unit mod. $9$. Indeed $6$ has only three multiples mod. $9$: $;6, 3$ and $0$.
$endgroup$
– Bernard
Mar 24 at 11:11
add a comment |
$begingroup$
Hint:
So you're given that $2$ has order $18bmod 19$. In particular, $7equiv 2^6mod 19$.
Any solution can be written as $2^k$, where $k$ is unique module $18$. The equation can be rewritten as
$$2^12kequiv 2^6 mod 19iff 12kequiv 6mod 18iff 2kequiv 1mod 3.$$
Can you continue?
answered Mar 23 at 19:56
BernardBernard
124k741117
$endgroup$
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input?
$endgroup$
– Kami
Mar 24 at 5:30
$begingroup$
@Kami: For the first equation, you find $kequiv 2mod 3$, i.e. $k=2, 5, 8$, &c. As to the second equation, $12kequiv 14pmod18iff 6kequiv 7pmod 9$, and there's not necessarily a solution, as $6$ is not a unit mod. $9$. Indeed $6$ has only three multiples mod. $9$: $;6, 3$ and $0$.
$endgroup$
– Bernard
Mar 24 at 11:11
add a comment |
$begingroup$
Hint:
So you're given that $2$ has order $18bmod 19$. In particular, $7equiv 2^6mod 19$.
Any solution can be written as $2^k$, where $k$ is unique module $18$. The equation can be rewritten as
$$2^12kequiv 2^6 mod 19iff 12kequiv 6mod 18iff 2kequiv 1mod 3.$$
Can you continue?
answered Mar 23 at 19:56
BernardBernard
124k741117
$endgroup$
Hint:
So you're given that $2$ has order $18bmod 19$. In particular, $7equiv 2^6mod 19$.
Any solution can be written as $2^k$, where $k$ is unique module $18$. The equation can be rewritten as
$$2^12kequiv 2^6 mod 19iff 12kequiv 6mod 18iff 2kequiv 1mod 3.$$
Can you continue?
answered Mar 23 at 19:56
BernardBernard
124k741117
answered Mar 23 at 19:56
BernardBernard
124k741117
answered Mar 23 at 19:56
BernardBernard
124k741117
answered Mar 23 at 19:56
BernardBernard
124k741117
124k741117
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input?
$endgroup$
– Kami
Mar 24 at 5:30
$begingroup$
@Kami: For the first equation, you find $kequiv 2mod 3$, i.e. $k=2, 5, 8$, &c. As to the second equation, $12kequiv 14pmod18iff 6kequiv 7pmod 9$, and there's not necessarily a solution, as $6$ is not a unit mod. $9$. Indeed $6$ has only three multiples mod. $9$: $;6, 3$ and $0$.
$endgroup$
– Bernard
Mar 24 at 11:11
add a comment |
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input?
$endgroup$
– Kami
Mar 24 at 5:30
$begingroup$
@Kami: For the first equation, you find $kequiv 2mod 3$, i.e. $k=2, 5, 8$, &c. As to the second equation, $12kequiv 14pmod18iff 6kequiv 7pmod 9$, and there's not necessarily a solution, as $6$ is not a unit mod. $9$. Indeed $6$ has only three multiples mod. $9$: $;6, 3$ and $0$.
$endgroup$
– Bernard
Mar 24 at 11:11
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input?
$endgroup$
– Kami
Mar 24 at 5:30
$begingroup$
This would solve to give k = 2, right? And then would the next one be 2^14 = 6 (mod 19) => 2^12k = 2^14 (mod 18)? I might have set this up wrong because I can't find a solution for 12k = 14 (mod 18). Any input?
$endgroup$
– Kami
Mar 24 at 5:30
$begingroup$
@Kami: For the first equation, you find $kequiv 2mod 3$, i.e. $k=2, 5, 8$, &c. As to the second equation, $12kequiv 14pmod18iff 6kequiv 7pmod 9$, and there's not necessarily a solution, as $6$ is not a unit mod. $9$. Indeed $6$ has only three multiples mod. $9$: $;6, 3$ and $0$.
$endgroup$
– Bernard
Mar 24 at 11:11
$begingroup$
@Kami: For the first equation, you find $kequiv 2mod 3$, i.e. $k=2, 5, 8$, &c. As to the second equation, $12kequiv 14pmod18iff 6kequiv 7pmod 9$, and there's not necessarily a solution, as $6$ is not a unit mod. $9$. Indeed $6$ has only three multiples mod. $9$: $;6, 3$ and $0$.
$endgroup$
– Bernard
Mar 24 at 11:11
add a comment |
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$begingroup$
You could start by listing out the powers of 2 $pmod19$...
$endgroup$
– Dzoooks
Mar 23 at 19:28