If $a^3 equiv b^3pmod11$ then $a equiv b pmod11$. [closed]
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$begingroup$
If $a^3 equiv b^3pmod11$ then $a equiv b pmod11$.
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
38.8k543153
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
$endgroup$
closed as off-topic by user21820, Did, amWhy, 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, amWhy, Lee David Chung Lin, KReiser
add a comment |
$begingroup$
If $a^3 equiv b^3pmod11$ then $a equiv b pmod11$.
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
38.8k543153
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
$endgroup$
closed as off-topic by user21820, Did, amWhy, 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, amWhy, Lee David Chung Lin, KReiser
add a comment |
$begingroup$
If $a^3 equiv b^3pmod11$ then $a equiv b pmod11$.
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
38.8k543153
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
$endgroup$
If $a^3 equiv b^3pmod11$ then $a equiv b pmod11$.
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
38.8k543153
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
edited Jan 7 at 14:12
user21820
38.8k543153
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
edited Jan 7 at 14:12
user21820
38.8k543153
edited Jan 7 at 14:12
user21820
38.8k543153
edited Jan 7 at 14:12
user21820
38.8k543153
38.8k543153
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
asked Jan 7 at 10:07
prashant sharmaprashant sharma
756
756
closed as off-topic by user21820, Did, amWhy, 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, amWhy, Lee David Chung Lin, KReiser
closed as off-topic by user21820, Did, amWhy, 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, amWhy, Lee David Chung Lin, KReiser
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
163k10169393
$endgroup$
$begingroup$
@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
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@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 |
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If $11mid ab$ we are done.
Say $11notmid ab$. Then by Fermat we have $$a^10equiv b^10equiv 1pmod 11$$
Since $$a^9equiv b^9pmod 11$$ we have $$a^10equiv ba^9pmod 11$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
$endgroup$
add a comment |
$begingroup$
Note $$1^3 equiv 2^3pmod7$$ doesnt imply $1equiv 2pmod7$
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
$endgroup$
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– greedoid
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
163k10169393
$endgroup$
$begingroup$
@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
163k10169393
$endgroup$
$begingroup$
@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
163k10169393
$endgroup$
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
163k10169393
edited Jan 7 at 22:58
answered Jan 7 at 10:27
lhflhf
163k10169393
answered Jan 7 at 10:27
lhflhf
163k10169393
answered Jan 7 at 10:27
lhflhf
163k10169393
163k10169393
$begingroup$
@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 $11notmid ab$. Then by Fermat we have $$a^10equiv b^10equiv 1pmod 11$$
Since $$a^9equiv b^9pmod 11$$ we have $$a^10equiv ba^9pmod 11$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
$endgroup$
add a comment |
$begingroup$
If $11mid ab$ we are done.
Say $11notmid ab$. Then by Fermat we have $$a^10equiv b^10equiv 1pmod 11$$
Since $$a^9equiv b^9pmod 11$$ we have $$a^10equiv ba^9pmod 11$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
$endgroup$
add a comment |
$begingroup$
If $11mid ab$ we are done.
Say $11notmid ab$. Then by Fermat we have $$a^10equiv b^10equiv 1pmod 11$$
Since $$a^9equiv b^9pmod 11$$ we have $$a^10equiv ba^9pmod 11$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
$endgroup$
If $11mid ab$ we are done.
Say $11notmid ab$. Then by Fermat we have $$a^10equiv b^10equiv 1pmod 11$$
Since $$a^9equiv b^9pmod 11$$ we have $$a^10equiv ba^9pmod 11$$
so $$ 11mid a^9(a-b)implies 11mid a-b$$
and we are done.
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
answered Jan 7 at 10:25
greedoidgreedoid
39.2k114797
39.2k114797
add a comment |
add a comment |
$begingroup$
Note $$1^3 equiv 2^3pmod7$$ doesnt imply $1equiv 2pmod7$
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
$endgroup$
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– greedoid
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 $$1^3 equiv 2^3pmod7$$ doesnt imply $1equiv 2pmod7$
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
$endgroup$
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– greedoid
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 $$1^3 equiv 2^3pmod7$$ doesnt imply $1equiv 2pmod7$
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
$endgroup$
Note $$1^3 equiv 2^3pmod7$$ doesnt imply $1equiv 2pmod7$
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
answered Jan 7 at 10:13
crskhrcrskhr
3,873925
3,873925
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– greedoid
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.
$endgroup$
– greedoid
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.
$endgroup$
– greedoid
Jan 7 at 10:31
$begingroup$
I don't understand this note. It is more like a comment, not a solution.
$endgroup$
– greedoid
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 |