Nonprimitive root of $($mod $p^2)$ [closed]
$begingroup$
Let p be an odd prime number.
Consider the set $S$ of CRS of $($mod $p^2)$. Now consider the subset $T$ $in$ $S$ : $x_s$ $equiv 2 ($mod $p)$.
What is the element $x_T$ of T that is NOT a primitive root $($mod $p^2)$? What is the process that can lead to find this one (or maybe many) solutions to the question?
The original problem was presented with p = 101
Thanks
elementary-number-theory modular-arithmetic
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
$endgroup$
closed as off-topic by Saad, José Carlos Santos, John B, supinf, amWhy Nov 26 '18 at 16:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, John B, supinf, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let p be an odd prime number.
Consider the set $S$ of CRS of $($mod $p^2)$. Now consider the subset $T$ $in$ $S$ : $x_s$ $equiv 2 ($mod $p)$.
What is the element $x_T$ of T that is NOT a primitive root $($mod $p^2)$? What is the process that can lead to find this one (or maybe many) solutions to the question?
The original problem was presented with p = 101
Thanks
elementary-number-theory modular-arithmetic
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
$endgroup$
closed as off-topic by Saad, José Carlos Santos, John B, supinf, amWhy Nov 26 '18 at 16:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, John B, supinf, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
By math.stackexchange.com/questions/227199/…, either $2$ or $2+101$ is a primitive root $pmod{101^2}$
$endgroup$
– lab bhattacharjee
Nov 26 '18 at 14:16
add a comment |
$begingroup$
Let p be an odd prime number.
Consider the set $S$ of CRS of $($mod $p^2)$. Now consider the subset $T$ $in$ $S$ : $x_s$ $equiv 2 ($mod $p)$.
What is the element $x_T$ of T that is NOT a primitive root $($mod $p^2)$? What is the process that can lead to find this one (or maybe many) solutions to the question?
The original problem was presented with p = 101
Thanks
elementary-number-theory modular-arithmetic
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
$endgroup$
Let p be an odd prime number.
Consider the set $S$ of CRS of $($mod $p^2)$. Now consider the subset $T$ $in$ $S$ : $x_s$ $equiv 2 ($mod $p)$.
What is the element $x_T$ of T that is NOT a primitive root $($mod $p^2)$? What is the process that can lead to find this one (or maybe many) solutions to the question?
The original problem was presented with p = 101
Thanks
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
edited Nov 26 '18 at 17:32
Alessar
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
asked Nov 26 '18 at 13:40
AlessarAlessar
27115
27115
closed as off-topic by Saad, José Carlos Santos, John B, supinf, amWhy Nov 26 '18 at 16:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, John B, supinf, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, José Carlos Santos, John B, supinf, amWhy Nov 26 '18 at 16:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, José Carlos Santos, John B, supinf, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
By math.stackexchange.com/questions/227199/…, either $2$ or $2+101$ is a primitive root $pmod{101^2}$
$endgroup$
– lab bhattacharjee
Nov 26 '18 at 14:16
add a comment |
$begingroup$
By math.stackexchange.com/questions/227199/…, either $2$ or $2+101$ is a primitive root $pmod{101^2}$
$endgroup$
– lab bhattacharjee
Nov 26 '18 at 14:16
$begingroup$
By math.stackexchange.com/questions/227199/…, either $2$ or $2+101$ is a primitive root $pmod{101^2}$
$endgroup$
– lab bhattacharjee
Nov 26 '18 at 14:16
$begingroup$
By math.stackexchange.com/questions/227199/…, either $2$ or $2+101$ is a primitive root $pmod{101^2}$
$endgroup$
– lab bhattacharjee
Nov 26 '18 at 14:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You know that $2$ is a primitive root of $101^2$ and that $phi(101^2) = 101cdot 100.$ If one of the numbers $a$ in your set is not a primitive root then you must have $a^{100}equiv 1 pmod{101^2}.$ So you're looking for an element with order $100$.
You know $2^{101cdot 100} equiv 1 pmod{101^2}.$ So $2^{101}$ has order dividing $100$. And by Fermat's little theorem, it's also congruent to $2 pmod{101}$.
So the answer is $2^{101} equiv 8385 pmod{101^2}.$
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You know that $2$ is a primitive root of $101^2$ and that $phi(101^2) = 101cdot 100.$ If one of the numbers $a$ in your set is not a primitive root then you must have $a^{100}equiv 1 pmod{101^2}.$ So you're looking for an element with order $100$.
You know $2^{101cdot 100} equiv 1 pmod{101^2}.$ So $2^{101}$ has order dividing $100$. And by Fermat's little theorem, it's also congruent to $2 pmod{101}$.
So the answer is $2^{101} equiv 8385 pmod{101^2}.$
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
$endgroup$
add a comment |
$begingroup$
You know that $2$ is a primitive root of $101^2$ and that $phi(101^2) = 101cdot 100.$ If one of the numbers $a$ in your set is not a primitive root then you must have $a^{100}equiv 1 pmod{101^2}.$ So you're looking for an element with order $100$.
You know $2^{101cdot 100} equiv 1 pmod{101^2}.$ So $2^{101}$ has order dividing $100$. And by Fermat's little theorem, it's also congruent to $2 pmod{101}$.
So the answer is $2^{101} equiv 8385 pmod{101^2}.$
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
$endgroup$
add a comment |
$begingroup$
You know that $2$ is a primitive root of $101^2$ and that $phi(101^2) = 101cdot 100.$ If one of the numbers $a$ in your set is not a primitive root then you must have $a^{100}equiv 1 pmod{101^2}.$ So you're looking for an element with order $100$.
You know $2^{101cdot 100} equiv 1 pmod{101^2}.$ So $2^{101}$ has order dividing $100$. And by Fermat's little theorem, it's also congruent to $2 pmod{101}$.
So the answer is $2^{101} equiv 8385 pmod{101^2}.$
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
$endgroup$
You know that $2$ is a primitive root of $101^2$ and that $phi(101^2) = 101cdot 100.$ If one of the numbers $a$ in your set is not a primitive root then you must have $a^{100}equiv 1 pmod{101^2}.$ So you're looking for an element with order $100$.
You know $2^{101cdot 100} equiv 1 pmod{101^2}.$ So $2^{101}$ has order dividing $100$. And by Fermat's little theorem, it's also congruent to $2 pmod{101}$.
So the answer is $2^{101} equiv 8385 pmod{101^2}.$
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
edited Nov 26 '18 at 15:26
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
answered Nov 26 '18 at 14:09
B. GoddardB. Goddard
18.6k21440
18.6k21440
add a comment |
add a comment |
$begingroup$
By math.stackexchange.com/questions/227199/…, either $2$ or $2+101$ is a primitive root $pmod{101^2}$
$endgroup$
– lab bhattacharjee
Nov 26 '18 at 14:16