If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$. [closed]
$begingroup$
If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.
I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?
elementary-number-theory modular-arithmetic
edited Jan 7 at 14:12
user21820
39.8k544158
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
$endgroup$
closed as off-topic by user21820, Did, Namaste, Lee David Chung Lin, KReiser Jan 8 at 1:36
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, Did, Namaste, Lee David Chung Lin, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.
I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?
elementary-number-theory modular-arithmetic
edited Jan 7 at 14:12
user21820
39.8k544158
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
$endgroup$
closed as off-topic by user21820, Did, Namaste, Lee David Chung Lin, KReiser Jan 8 at 1:36
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, Did, Namaste, Lee David Chung Lin, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.
I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?
elementary-number-theory modular-arithmetic
edited Jan 7 at 14:12
user21820
39.8k544158
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
$endgroup$
If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$.
I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
edited Jan 7 at 14:12
user21820
39.8k544158
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
edited Jan 7 at 14:12
user21820
39.8k544158
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
edited Jan 7 at 14:12
user21820
39.8k544158
edited Jan 7 at 14:12
user21820
39.8k544158
edited Jan 7 at 14:12
user21820
39.8k544158
39.8k544158
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
asked Jan 7 at 10:07
prashant sharmaprashant sharma
827
827
closed as off-topic by user21820, Did, Namaste, Lee David Chung Lin, KReiser Jan 8 at 1:36
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, Did, Namaste, Lee David Chung Lin, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Did, Namaste, Lee David Chung Lin, KReiser Jan 8 at 1:36
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, Did, Namaste, Lee David Chung Lin, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is a general statement:
If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.
This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.
In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
$$
a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
equiv
b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
bmod 11
$$
answered Jan 7 at 10:27
lhflhf
167k11172403
$endgroup$
$begingroup$
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
$endgroup$
– prashant sharma
Jan 7 at 10:33
$begingroup$
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
$endgroup$
– lhf
Jan 7 at 10:35
$begingroup$
@Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
$endgroup$
– prashant sharma
Jan 7 at 16:13
$begingroup$
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
$endgroup$
– lhf
Jan 7 at 22:58
add a comment |
$begingroup$
If $11mid ab$ we are done.
Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$
Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
add a comment |
$begingroup$
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
$endgroup$
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– Maria Mazur
Jan 7 at 10:31
$begingroup$
"Can we replace 11 by any other integer" is what i have answered
$endgroup$
– crskhr
Jan 7 at 10:33
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a general statement:
If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.
This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.
In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
$$
a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
equiv
b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
bmod 11
$$
answered Jan 7 at 10:27
lhflhf
167k11172403
$endgroup$
$begingroup$
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
$endgroup$
– prashant sharma
Jan 7 at 10:33
$begingroup$
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
$endgroup$
– lhf
Jan 7 at 10:35
$begingroup$
@Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
$endgroup$
– prashant sharma
Jan 7 at 16:13
$begingroup$
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
$endgroup$
– lhf
Jan 7 at 22:58
add a comment |
$begingroup$
Here is a general statement:
If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.
This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.
In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
$$
a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
equiv
b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
bmod 11
$$
answered Jan 7 at 10:27
lhflhf
167k11172403
$endgroup$
$begingroup$
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
$endgroup$
– prashant sharma
Jan 7 at 10:33
$begingroup$
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
$endgroup$
– lhf
Jan 7 at 10:35
$begingroup$
@Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
$endgroup$
– prashant sharma
Jan 7 at 16:13
$begingroup$
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
$endgroup$
– lhf
Jan 7 at 22:58
add a comment |
$begingroup$
Here is a general statement:
If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.
This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.
In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
$$
a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
equiv
b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
bmod 11
$$
answered Jan 7 at 10:27
lhflhf
167k11172403
$endgroup$
Here is a general statement:
If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m equiv b^m bmod p$ iff $a equiv bbmod p$.
This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v in mathbb N$.
In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10cdot (-2) + 3cdot(7)=1$. Write this as $1 + 10 cdot 2 = 3cdot 7 $. Then
$$
a equiv a cdot a^{10 cdot 2} = a^{1 + 10 cdot 2} = a^{3cdot7}
equiv
b^{3cdot7} = b^{1 + 10 cdot 2} = b cdot b^{10 cdot 2} equiv b
bmod 11
$$
answered Jan 7 at 10:27
lhflhf
167k11172403
edited Jan 7 at 22:58
answered Jan 7 at 10:27
lhflhf
167k11172403
answered Jan 7 at 10:27
lhflhf
167k11172403
answered Jan 7 at 10:27
lhflhf
167k11172403
167k11172403
$begingroup$
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
$endgroup$
– prashant sharma
Jan 7 at 10:33
$begingroup$
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
$endgroup$
– lhf
Jan 7 at 10:35
$begingroup$
@Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
$endgroup$
– prashant sharma
Jan 7 at 16:13
$begingroup$
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
$endgroup$
– lhf
Jan 7 at 22:58
add a comment |
$begingroup$
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
$endgroup$
– prashant sharma
Jan 7 at 10:33
$begingroup$
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
$endgroup$
– lhf
Jan 7 at 10:35
$begingroup$
@Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
$endgroup$
– prashant sharma
Jan 7 at 16:13
$begingroup$
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
$endgroup$
– lhf
Jan 7 at 22:58
$begingroup$
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
$endgroup$
– prashant sharma
Jan 7 at 10:33
$begingroup$
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
$endgroup$
– prashant sharma
Jan 7 at 10:33
$begingroup$
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
$endgroup$
– lhf
Jan 7 at 10:35
$begingroup$
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
$endgroup$
– lhf
Jan 7 at 10:35
$begingroup$
@Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
$endgroup$
– prashant sharma
Jan 7 at 16:13
$begingroup$
@Ihf Let's take $m = 3$, $p = 11$. Then one possible ordered pair of $(u, v)$ is $(1, -3)$. So we have $a^{10} equiv b^{10}$(mod $11$) and $a^{-9} equiv b^{-9}$(mod $11$) and we multiply both congruences then we get $a equiv b$(mod $11$). But second congruence is meaning less as exponents are negative.
$endgroup$
– prashant sharma
Jan 7 at 16:13
$begingroup$
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
$endgroup$
– lhf
Jan 7 at 22:58
$begingroup$
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
$endgroup$
– lhf
Jan 7 at 22:58
add a comment |
$begingroup$
If $11mid ab$ we are done.
Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$
Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
add a comment |
$begingroup$
If $11mid ab$ we are done.
Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$
Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
add a comment |
$begingroup$
If $11mid ab$ we are done.
Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$
Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
If $11mid ab$ we are done.
Say $11not{mid} ab$. Then by Fermat we have $$a^{10}equiv b^{10}equiv 1pmod {11}$$
Since $$a^{9}equiv b^{9}pmod {11}$$ we have $$a^{10}equiv ba^{9}pmod {11}$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
answered Jan 7 at 10:25
Maria MazurMaria Mazur
48.5k1260121
48.5k1260121
add a comment |
add a comment |
$begingroup$
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
$endgroup$
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– Maria Mazur
Jan 7 at 10:31
$begingroup$
"Can we replace 11 by any other integer" is what i have answered
$endgroup$
– crskhr
Jan 7 at 10:33
add a comment |
$begingroup$
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
$endgroup$
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– Maria Mazur
Jan 7 at 10:31
$begingroup$
"Can we replace 11 by any other integer" is what i have answered
$endgroup$
– crskhr
Jan 7 at 10:33
add a comment |
$begingroup$
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
$endgroup$
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
answered Jan 7 at 10:13
crskhrcrskhr
3,890926
3,890926
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– Maria Mazur
Jan 7 at 10:31
$begingroup$
"Can we replace 11 by any other integer" is what i have answered
$endgroup$
– crskhr
Jan 7 at 10:33
add a comment |
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– Maria Mazur
Jan 7 at 10:31
$begingroup$
"Can we replace 11 by any other integer" is what i have answered
$endgroup$
– crskhr
Jan 7 at 10:33
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– Maria Mazur
Jan 7 at 10:31
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– Maria Mazur
Jan 7 at 10:31
$begingroup$
"Can we replace 11 by any other integer" is what i have answered
$endgroup$
– crskhr
Jan 7 at 10:33
$begingroup$
"Can we replace 11 by any other integer" is what i have answered
$endgroup$
– crskhr
Jan 7 at 10:33
add a comment |
