If $a^3 equiv b^3pmod{11}$ then $a equiv b pmod{11}$. [on hold]
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 yesterday
user21820
38.7k543153
asked yesterday
prashant sharmaprashant sharma
756
put on hold as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser 21 hours ago
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, amWhy, 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 |
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 yesterday
user21820
38.7k543153
asked yesterday
prashant sharmaprashant sharma
756
put on hold as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser 21 hours ago
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, amWhy, 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 |
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 yesterday
user21820
38.7k543153
asked yesterday
prashant sharmaprashant sharma
756
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 yesterday
user21820
38.7k543153
asked yesterday
prashant sharmaprashant sharma
756
edited yesterday
user21820
38.7k543153
asked yesterday
prashant sharmaprashant sharma
756
edited yesterday
user21820
38.7k543153
edited yesterday
user21820
38.7k543153
edited yesterday
user21820
38.7k543153
38.7k543153
asked yesterday
prashant sharmaprashant sharma
756
asked yesterday
prashant sharmaprashant sharma
756
asked yesterday
prashant sharmaprashant sharma
756
756
put on hold as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser 21 hours ago
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, amWhy, Lee David Chung Lin, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user21820, Did, amWhy, Lee David Chung Lin, KReiser 21 hours ago
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, amWhy, 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
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 yesterday
lhflhf
163k10167388
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
– prashant sharma
yesterday
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
– lhf
yesterday
@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.
– prashant sharma
yesterday
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
– lhf
yesterday
add a comment |
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 yesterday
greedoidgreedoid
38.5k114797
add a comment |
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered yesterday
crskhrcrskhr
3,873925
I don't understand this note. It is more like a comment, not a solution.
– greedoid
yesterday
"Can we replace 11 by any other integer" is what i have answered
– crskhr
yesterday
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 yesterday
lhflhf
163k10167388
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
– prashant sharma
yesterday
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
– lhf
yesterday
@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.
– prashant sharma
yesterday
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
– lhf
yesterday
add a comment |
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 yesterday
lhflhf
163k10167388
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
– prashant sharma
yesterday
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
– lhf
yesterday
@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.
– prashant sharma
yesterday
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
– lhf
yesterday
add a comment |
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 yesterday
lhflhf
163k10167388
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 yesterday
lhflhf
163k10167388
edited yesterday
answered yesterday
lhflhf
163k10167388
answered yesterday
lhflhf
163k10167388
answered yesterday
lhflhf
163k10167388
163k10167388
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
– prashant sharma
yesterday
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
– lhf
yesterday
@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.
– prashant sharma
yesterday
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
– lhf
yesterday
add a comment |
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
– prashant sharma
yesterday
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
– lhf
yesterday
@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.
– prashant sharma
yesterday
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
– lhf
yesterday
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
– prashant sharma
yesterday
@Ihf I think that above argument would work if both $u$ and $v$ are positive integers but it is not the case here.
– prashant sharma
yesterday
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
– lhf
yesterday
@prashantsharma, exactly one of $u,v$ is negative. Move the corresponding therm to the other side.
– lhf
yesterday
@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.
– prashant sharma
yesterday
@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.
– prashant sharma
yesterday
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
– lhf
yesterday
@prashantsharma, negative exponents do make sense but anyway see my edited answer.
– lhf
yesterday
add a comment |
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 yesterday
greedoidgreedoid
38.5k114797
add a comment |
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 yesterday
greedoidgreedoid
38.5k114797
add a comment |
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 yesterday
greedoidgreedoid
38.5k114797
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 yesterday
greedoidgreedoid
38.5k114797
answered yesterday
greedoidgreedoid
38.5k114797
answered yesterday
greedoidgreedoid
38.5k114797
answered yesterday
greedoidgreedoid
38.5k114797
38.5k114797
add a comment |
add a comment |
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered yesterday
crskhrcrskhr
3,873925
I don't understand this note. It is more like a comment, not a solution.
– greedoid
yesterday
"Can we replace 11 by any other integer" is what i have answered
– crskhr
yesterday
add a comment |
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered yesterday
crskhrcrskhr
3,873925
I don't understand this note. It is more like a comment, not a solution.
– greedoid
yesterday
"Can we replace 11 by any other integer" is what i have answered
– crskhr
yesterday
add a comment |
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered yesterday
crskhrcrskhr
3,873925
Note $$1^{3} equiv 2^{3}pmod{7}$$ doesnt imply $1equiv 2pmod{7}$
answered yesterday
crskhrcrskhr
3,873925
answered yesterday
crskhrcrskhr
3,873925
answered yesterday
crskhrcrskhr
3,873925
answered yesterday
crskhrcrskhr
3,873925
3,873925
I don't understand this note. It is more like a comment, not a solution.
– greedoid
yesterday
"Can we replace 11 by any other integer" is what i have answered
– crskhr
yesterday
add a comment |
I don't understand this note. It is more like a comment, not a solution.
– greedoid
yesterday
"Can we replace 11 by any other integer" is what i have answered
– crskhr
yesterday
I don't understand this note. It is more like a comment, not a solution.
– greedoid
yesterday
I don't understand this note. It is more like a comment, not a solution.
– greedoid
yesterday
"Can we replace 11 by any other integer" is what i have answered
– crskhr
yesterday
"Can we replace 11 by any other integer" is what i have answered
– crskhr
yesterday
add a comment |
