True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$. [closed]
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True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.
Do I use Fermat's Little Theorem or its corollary?
modular-arithmetic
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
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closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.
Do I use Fermat's Little Theorem or its corollary?
modular-arithmetic
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
$endgroup$
closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42
1
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Have you learnt about the Legendre symbol yet?
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– Lord Shark the Unknown
Dec 9 '18 at 4:43
1
$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55
add a comment |
$begingroup$
True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.
Do I use Fermat's Little Theorem or its corollary?
modular-arithmetic
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
$endgroup$
True or false, there exist exactly two elements $[x]$ in $mathbb{Z}_{11}$ such that $[x]^2 = [2]$.
Do I use Fermat's Little Theorem or its corollary?
modular-arithmetic
modular-arithmetic
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
asked Dec 9 '18 at 4:38
hunnybunshunnybuns
15
15
closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, user10354138, José Carlos Santos, Gibbs, John B Dec 10 '18 at 11:40
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, José Carlos Santos, Gibbs, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42
1
$begingroup$
Have you learnt about the Legendre symbol yet?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 4:43
1
$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55
add a comment |
2
$begingroup$
Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42
1
$begingroup$
Have you learnt about the Legendre symbol yet?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 4:43
1
$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55
2
2
$begingroup$
Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42
$begingroup$
Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42
1
1
$begingroup$
Have you learnt about the Legendre symbol yet?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 4:43
$begingroup$
Have you learnt about the Legendre symbol yet?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 4:43
1
1
$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55
$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Checking one at a time is one approach. It's false.
Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
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add a comment |
$begingroup$
Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
begin{align*}
x^2 & equiv 2 pmod{11}\
2^{2k} & equiv 2 pmod{11}\
2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
end{align*}
But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Checking one at a time is one approach. It's false.
Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
$endgroup$
add a comment |
$begingroup$
Checking one at a time is one approach. It's false.
Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
$endgroup$
add a comment |
$begingroup$
Checking one at a time is one approach. It's false.
Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
$endgroup$
Checking one at a time is one approach. It's false.
Now using Fermat's little theorem, if $x^2cong2pmod{11}$, it follows that $(x^2)^5cong32pmod{11}cong10pmod{11}$. So again we get that it's false: actually there are no such elements.
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
edited Dec 9 '18 at 4:55
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
answered Dec 9 '18 at 4:47
Chris CusterChris Custer
11.6k3824
11.6k3824
add a comment |
add a comment |
$begingroup$
Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
begin{align*}
x^2 & equiv 2 pmod{11}\
2^{2k} & equiv 2 pmod{11}\
2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
end{align*}
But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
$endgroup$
add a comment |
$begingroup$
Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
begin{align*}
x^2 & equiv 2 pmod{11}\
2^{2k} & equiv 2 pmod{11}\
2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
end{align*}
But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
$endgroup$
add a comment |
$begingroup$
Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
begin{align*}
x^2 & equiv 2 pmod{11}\
2^{2k} & equiv 2 pmod{11}\
2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
end{align*}
But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
$endgroup$
Another approach: Observe that every non-zero element of $Bbb{Z}_{11}$ can be written as a power of $2$ (same as saying $2$ is a primitive root or a generator). If such an $x$ exists, then $x=2^k$ for some $k in {1,2, ldots ,10 }$. This implies:
begin{align*}
x^2 & equiv 2 pmod{11}\
2^{2k} & equiv 2 pmod{11}\
2^{2k-1} & equiv 1 pmod{11} & (because gcd(2,11)=1)
end{align*}
But then $10$ must divide $2k-1$, which is impossible. So no such $x$ exists.
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
answered Dec 9 '18 at 6:00
Anurag AAnurag A
26k12249
26k12249
add a comment |
add a comment |
2
$begingroup$
Only 11? Just check them all
$endgroup$
– Randall
Dec 9 '18 at 4:42
1
$begingroup$
Have you learnt about the Legendre symbol yet?
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 4:43
1
$begingroup$
You only need to check half since $x^2 = (-x)^2$
$endgroup$
– qwr
Dec 9 '18 at 4:55