Say if there is an element in $Bbb Z_{900}$ with 30 as additive and multiplicative order The...
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Say if there is an element in $Bbb Z_{900}$ with 30 as additive and multiplicative order
The Next CEO of Stack OverflowProve that there are $736$ $2 times 2$ matrices ($A$) where $A=A^{-1}$Proof that $mathbb Z[sqrt{3}]$ is a Euclidean DomainIf there is an $ainmathbb{Z}$ with $a^{n-1}equiv 1mod n$ but $a^{frac{n-1}p}notequiv 1$ for all primes $pmid n-1$, then $n$ is a primeSuppose that $R$ is a commutative ring with no zero divisors. Show that all non zero elements of $R$ have the same additive order.Does a system om congruence equations have solutions?Proof solutions linear congruenceSolving a linear system in $mathbb{Z}_{12}$?It is possible to show/prove that the cancellation property is necessary to prove $0x=0$ for $xnotin mathbb{Z}^+ cup {0}$?Assume $p$ is prime, $axequiv 0 (mod p)$ and $pnmid a$, then the additive order of $a$ is $p$ $(mod p$.Solve a system of congruences using the Chinese Remainder Theorem
$begingroup$
I have a ring $Bbb Z_{900}$ and I need to check if there is an element which has multiplicative and additive order 30.
I think that is sufficient to check if the following system has at least one solution:
$$begin{cases}
begin{aligned}
30x&equiv0 (text{mod} 900) \
x^{30}&equiv1 (text{mod} 900)
end{aligned}
end{cases}$$
If my attempt is correct, how can I solve the system? The first congruence is easy to solve. How can I solve the second? Is there a method to reduce the power?
abstract-algebra discrete-mathematics ring-theory
edited Mar 16 at 18:18
MarianD
1,7791617
asked Mar 16 at 18:06
JackJack
877
$endgroup$
|
show 3 more comments
$begingroup$
I have a ring $Bbb Z_{900}$ and I need to check if there is an element which has multiplicative and additive order 30.
I think that is sufficient to check if the following system has at least one solution:
$$begin{cases}
begin{aligned}
30x&equiv0 (text{mod} 900) \
x^{30}&equiv1 (text{mod} 900)
end{aligned}
end{cases}$$
If my attempt is correct, how can I solve the system? The first congruence is easy to solve. How can I solve the second? Is there a method to reduce the power?
abstract-algebra discrete-mathematics ring-theory
edited Mar 16 at 18:18
MarianD
1,7791617
asked Mar 16 at 18:06
JackJack
877
$endgroup$
$begingroup$
Hi Jack. What numbers can satisfy the first equation in your system? --- Consider that first, then see if there is any way any of them can satisfy the second. It might be useful to use that 900=30*30
$endgroup$
– kholli
Mar 16 at 18:12
$begingroup$
the first equation is true for $x=900t, t in Z$
$endgroup$
– Jack
Mar 16 at 18:15
$begingroup$
Is it true for 30? Is $30 = 900t$ for some $tin mathbb{Z}$? You might want to put some more thought into the numbers that satisfy equation 1.
$endgroup$
– kholli
Mar 16 at 18:16
$begingroup$
could you be more detailed? I don't think I understand what you mean.
$endgroup$
– Jack
Mar 16 at 18:20
1
$begingroup$
Any element satisfying the second equality will be invertible. But then it will have order what additively?
$endgroup$
– Tobias Kildetoft
Mar 16 at 18:24
|
show 3 more comments
$begingroup$
I have a ring $Bbb Z_{900}$ and I need to check if there is an element which has multiplicative and additive order 30.
I think that is sufficient to check if the following system has at least one solution:
$$begin{cases}
begin{aligned}
30x&equiv0 (text{mod} 900) \
x^{30}&equiv1 (text{mod} 900)
end{aligned}
end{cases}$$
If my attempt is correct, how can I solve the system? The first congruence is easy to solve. How can I solve the second? Is there a method to reduce the power?
abstract-algebra discrete-mathematics ring-theory
edited Mar 16 at 18:18
MarianD
1,7791617
asked Mar 16 at 18:06
JackJack
877
$endgroup$
I have a ring $Bbb Z_{900}$ and I need to check if there is an element which has multiplicative and additive order 30.
I think that is sufficient to check if the following system has at least one solution:
$$begin{cases}
begin{aligned}
30x&equiv0 (text{mod} 900) \
x^{30}&equiv1 (text{mod} 900)
end{aligned}
end{cases}$$
If my attempt is correct, how can I solve the system? The first congruence is easy to solve. How can I solve the second? Is there a method to reduce the power?
abstract-algebra discrete-mathematics ring-theory
abstract-algebra discrete-mathematics ring-theory
edited Mar 16 at 18:18
MarianD
1,7791617
asked Mar 16 at 18:06
JackJack
877
edited Mar 16 at 18:18
MarianD
1,7791617
asked Mar 16 at 18:06
JackJack
877
edited Mar 16 at 18:18
MarianD
1,7791617
edited Mar 16 at 18:18
MarianD
1,7791617
edited Mar 16 at 18:18
MarianD
1,7791617
1,7791617
asked Mar 16 at 18:06
JackJack
877
asked Mar 16 at 18:06
JackJack
877
asked Mar 16 at 18:06
JackJack
877
877
$begingroup$
Hi Jack. What numbers can satisfy the first equation in your system? --- Consider that first, then see if there is any way any of them can satisfy the second. It might be useful to use that 900=30*30
$endgroup$
– kholli
Mar 16 at 18:12
$begingroup$
the first equation is true for $x=900t, t in Z$
$endgroup$
– Jack
Mar 16 at 18:15
$begingroup$
Is it true for 30? Is $30 = 900t$ for some $tin mathbb{Z}$? You might want to put some more thought into the numbers that satisfy equation 1.
$endgroup$
– kholli
Mar 16 at 18:16
$begingroup$
could you be more detailed? I don't think I understand what you mean.
$endgroup$
– Jack
Mar 16 at 18:20
1
$begingroup$
Any element satisfying the second equality will be invertible. But then it will have order what additively?
$endgroup$
– Tobias Kildetoft
Mar 16 at 18:24
|
show 3 more comments
$begingroup$
Hi Jack. What numbers can satisfy the first equation in your system? --- Consider that first, then see if there is any way any of them can satisfy the second. It might be useful to use that 900=30*30
$endgroup$
– kholli
Mar 16 at 18:12
$begingroup$
the first equation is true for $x=900t, t in Z$
$endgroup$
– Jack
Mar 16 at 18:15
$begingroup$
Is it true for 30? Is $30 = 900t$ for some $tin mathbb{Z}$? You might want to put some more thought into the numbers that satisfy equation 1.
$endgroup$
– kholli
Mar 16 at 18:16
$begingroup$
could you be more detailed? I don't think I understand what you mean.
$endgroup$
– Jack
Mar 16 at 18:20
1
$begingroup$
Any element satisfying the second equality will be invertible. But then it will have order what additively?
$endgroup$
– Tobias Kildetoft
Mar 16 at 18:24
$begingroup$
Hi Jack. What numbers can satisfy the first equation in your system? --- Consider that first, then see if there is any way any of them can satisfy the second. It might be useful to use that 900=30*30
$endgroup$
– kholli
Mar 16 at 18:12
$begingroup$
Hi Jack. What numbers can satisfy the first equation in your system? --- Consider that first, then see if there is any way any of them can satisfy the second. It might be useful to use that 900=30*30
$endgroup$
– kholli
Mar 16 at 18:12
$begingroup$
the first equation is true for $x=900t, t in Z$
$endgroup$
– Jack
Mar 16 at 18:15
$begingroup$
the first equation is true for $x=900t, t in Z$
$endgroup$
– Jack
Mar 16 at 18:15
$begingroup$
Is it true for 30? Is $30 = 900t$ for some $tin mathbb{Z}$? You might want to put some more thought into the numbers that satisfy equation 1.
$endgroup$
– kholli
Mar 16 at 18:16
$begingroup$
Is it true for 30? Is $30 = 900t$ for some $tin mathbb{Z}$? You might want to put some more thought into the numbers that satisfy equation 1.
$endgroup$
– kholli
Mar 16 at 18:16
$begingroup$
could you be more detailed? I don't think I understand what you mean.
$endgroup$
– Jack
Mar 16 at 18:20
$begingroup$
could you be more detailed? I don't think I understand what you mean.
$endgroup$
– Jack
Mar 16 at 18:20
1
1
$begingroup$
Any element satisfying the second equality will be invertible. But then it will have order what additively?
$endgroup$
– Tobias Kildetoft
Mar 16 at 18:24
$begingroup$
Any element satisfying the second equality will be invertible. But then it will have order what additively?
$endgroup$
– Tobias Kildetoft
Mar 16 at 18:24
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N} Rightarrow x = 30z$. Now then $(30z)^3 = 30^3z^3 = 0 , , (text{mod } 900)$. So there isn't any solution to that system.
answered Mar 16 at 20:12
NawajNawaj
534
$endgroup$
$begingroup$
Thank you for your answer, could you please be more detailed in the steps? For example, how do I get this step $30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N}$?
$endgroup$
– Jack
Mar 17 at 9:27
$begingroup$
$900/30x$ for $30x = 0$ mod 900. $z in mathbb{Z}$ but since we are talking about elements in $Z_{900}$, we can only consider positive integers and including 0.
$endgroup$
– Nawaj
Mar 17 at 20:22
$begingroup$
Forgive me, but I still don't understand.I don't understand how you can write $30𝑥=900𝑧$ without first "removing" 30... I've never seen this way of resolving a linear congruence
$endgroup$
– Jack
Mar 19 at 16:10
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
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oldest
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$begingroup$
$30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N} Rightarrow x = 30z$. Now then $(30z)^3 = 30^3z^3 = 0 , , (text{mod } 900)$. So there isn't any solution to that system.
answered Mar 16 at 20:12
NawajNawaj
534
$endgroup$
$begingroup$
Thank you for your answer, could you please be more detailed in the steps? For example, how do I get this step $30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N}$?
$endgroup$
– Jack
Mar 17 at 9:27
$begingroup$
$900/30x$ for $30x = 0$ mod 900. $z in mathbb{Z}$ but since we are talking about elements in $Z_{900}$, we can only consider positive integers and including 0.
$endgroup$
– Nawaj
Mar 17 at 20:22
$begingroup$
Forgive me, but I still don't understand.I don't understand how you can write $30𝑥=900𝑧$ without first "removing" 30... I've never seen this way of resolving a linear congruence
$endgroup$
– Jack
Mar 19 at 16:10
add a comment |
$begingroup$
$30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N} Rightarrow x = 30z$. Now then $(30z)^3 = 30^3z^3 = 0 , , (text{mod } 900)$. So there isn't any solution to that system.
answered Mar 16 at 20:12
NawajNawaj
534
$endgroup$
$begingroup$
Thank you for your answer, could you please be more detailed in the steps? For example, how do I get this step $30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N}$?
$endgroup$
– Jack
Mar 17 at 9:27
$begingroup$
$900/30x$ for $30x = 0$ mod 900. $z in mathbb{Z}$ but since we are talking about elements in $Z_{900}$, we can only consider positive integers and including 0.
$endgroup$
– Nawaj
Mar 17 at 20:22
$begingroup$
Forgive me, but I still don't understand.I don't understand how you can write $30𝑥=900𝑧$ without first "removing" 30... I've never seen this way of resolving a linear congruence
$endgroup$
– Jack
Mar 19 at 16:10
add a comment |
$begingroup$
$30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N} Rightarrow x = 30z$. Now then $(30z)^3 = 30^3z^3 = 0 , , (text{mod } 900)$. So there isn't any solution to that system.
answered Mar 16 at 20:12
NawajNawaj
534
$endgroup$
$30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N} Rightarrow x = 30z$. Now then $(30z)^3 = 30^3z^3 = 0 , , (text{mod } 900)$. So there isn't any solution to that system.
answered Mar 16 at 20:12
NawajNawaj
534
answered Mar 16 at 20:12
NawajNawaj
534
answered Mar 16 at 20:12
NawajNawaj
534
answered Mar 16 at 20:12
NawajNawaj
534
534
$begingroup$
Thank you for your answer, could you please be more detailed in the steps? For example, how do I get this step $30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N}$?
$endgroup$
– Jack
Mar 17 at 9:27
$begingroup$
$900/30x$ for $30x = 0$ mod 900. $z in mathbb{Z}$ but since we are talking about elements in $Z_{900}$, we can only consider positive integers and including 0.
$endgroup$
– Nawaj
Mar 17 at 20:22
$begingroup$
Forgive me, but I still don't understand.I don't understand how you can write $30𝑥=900𝑧$ without first "removing" 30... I've never seen this way of resolving a linear congruence
$endgroup$
– Jack
Mar 19 at 16:10
add a comment |
$begingroup$
Thank you for your answer, could you please be more detailed in the steps? For example, how do I get this step $30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N}$?
$endgroup$
– Jack
Mar 17 at 9:27
$begingroup$
$900/30x$ for $30x = 0$ mod 900. $z in mathbb{Z}$ but since we are talking about elements in $Z_{900}$, we can only consider positive integers and including 0.
$endgroup$
– Nawaj
Mar 17 at 20:22
$begingroup$
Forgive me, but I still don't understand.I don't understand how you can write $30𝑥=900𝑧$ without first "removing" 30... I've never seen this way of resolving a linear congruence
$endgroup$
– Jack
Mar 19 at 16:10
$begingroup$
Thank you for your answer, could you please be more detailed in the steps? For example, how do I get this step $30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N}$?
$endgroup$
– Jack
Mar 17 at 9:27
$begingroup$
Thank you for your answer, could you please be more detailed in the steps? For example, how do I get this step $30x equiv 0 , , (text{mod } 900) Rightarrow 30x = 900z ni z in mathbb{N}$?
$endgroup$
– Jack
Mar 17 at 9:27
$begingroup$
$900/30x$ for $30x = 0$ mod 900. $z in mathbb{Z}$ but since we are talking about elements in $Z_{900}$, we can only consider positive integers and including 0.
$endgroup$
– Nawaj
Mar 17 at 20:22
$begingroup$
$900/30x$ for $30x = 0$ mod 900. $z in mathbb{Z}$ but since we are talking about elements in $Z_{900}$, we can only consider positive integers and including 0.
$endgroup$
– Nawaj
Mar 17 at 20:22
$begingroup$
Forgive me, but I still don't understand.I don't understand how you can write $30𝑥=900𝑧$ without first "removing" 30... I've never seen this way of resolving a linear congruence
$endgroup$
– Jack
Mar 19 at 16:10
$begingroup$
Forgive me, but I still don't understand.I don't understand how you can write $30𝑥=900𝑧$ without first "removing" 30... I've never seen this way of resolving a linear congruence
$endgroup$
– Jack
Mar 19 at 16:10
add a comment |
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$begingroup$
Hi Jack. What numbers can satisfy the first equation in your system? --- Consider that first, then see if there is any way any of them can satisfy the second. It might be useful to use that 900=30*30
$endgroup$
– kholli
Mar 16 at 18:12
$begingroup$
the first equation is true for $x=900t, t in Z$
$endgroup$
– Jack
Mar 16 at 18:15
$begingroup$
Is it true for 30? Is $30 = 900t$ for some $tin mathbb{Z}$? You might want to put some more thought into the numbers that satisfy equation 1.
$endgroup$
– kholli
Mar 16 at 18:16
$begingroup$
could you be more detailed? I don't think I understand what you mean.
$endgroup$
– Jack
Mar 16 at 18:20
1
$begingroup$
Any element satisfying the second equality will be invertible. But then it will have order what additively?
$endgroup$
– Tobias Kildetoft
Mar 16 at 18:24