Mixing
Prove that if $p$ is prime and $a^p+b^p = c^p$ then $a+b-c = 0 mod p$
$begingroup$
I'm working on the following Fermat little theorem exercise:
Prove that if $p$ is prime and $a^p+b^p = c^p$ then $a+b-c = 0 mod p$
Also I find a relation with Fermat last theorem which says that no three positive integers $a, b$, and $c$ satisfy the equation $a^n$ + $b^n$ $=$ $c^n$ for any integer value of $n$ greater than $2$.
So is there a solution or a way to solve the problem based on the last theorem? How should I go ahead on this exercise? Any hint or help will be really appreciated.
number-theory modular-arithmetic
asked Jan 29 at 6:39
mrazmraz
44319
$endgroup$
add a comment |
$begingroup$
I'm working on the following Fermat little theorem exercise:
Prove that if $p$ is prime and $a^p+b^p = c^p$ then $a+b-c = 0 mod p$
Also I find a relation with Fermat last theorem which says that no three positive integers $a, b$, and $c$ satisfy the equation $a^n$ + $b^n$ $=$ $c^n$ for any integer value of $n$ greater than $2$.
So is there a solution or a way to solve the problem based on the last theorem? How should I go ahead on this exercise? Any hint or help will be really appreciated.
number-theory modular-arithmetic
asked Jan 29 at 6:39
mrazmraz
44319
$endgroup$
$begingroup$
Fermat's Last Theorem is a distraction here - it says, amongst other things, that there are no solutions to the original for $pgt 2$. A false statement implies anything, so the implication is then trivially true for $pgt 2$ and easy for $p=2$. But really the statement here is an early and elementary observation in the journey towards proving the last theorem.
$endgroup$
– Mark Bennet
Jan 29 at 7:43
add a comment |
$begingroup$
I'm working on the following Fermat little theorem exercise:
Prove that if $p$ is prime and $a^p+b^p = c^p$ then $a+b-c = 0 mod p$
Also I find a relation with Fermat last theorem which says that no three positive integers $a, b$, and $c$ satisfy the equation $a^n$ + $b^n$ $=$ $c^n$ for any integer value of $n$ greater than $2$.
So is there a solution or a way to solve the problem based on the last theorem? How should I go ahead on this exercise? Any hint or help will be really appreciated.
number-theory modular-arithmetic
asked Jan 29 at 6:39
mrazmraz
44319
$endgroup$
I'm working on the following Fermat little theorem exercise:
Prove that if $p$ is prime and $a^p+b^p = c^p$ then $a+b-c = 0 mod p$
Also I find a relation with Fermat last theorem which says that no three positive integers $a, b$, and $c$ satisfy the equation $a^n$ + $b^n$ $=$ $c^n$ for any integer value of $n$ greater than $2$.
So is there a solution or a way to solve the problem based on the last theorem? How should I go ahead on this exercise? Any hint or help will be really appreciated.
number-theory modular-arithmetic
number-theory modular-arithmetic
asked Jan 29 at 6:39
mrazmraz
44319
asked Jan 29 at 6:39
mrazmraz
44319
asked Jan 29 at 6:39
mrazmraz
44319
asked Jan 29 at 6:39
mrazmraz
44319
asked Jan 29 at 6:39
mrazmraz
44319
44319
$begingroup$
Fermat's Last Theorem is a distraction here - it says, amongst other things, that there are no solutions to the original for $pgt 2$. A false statement implies anything, so the implication is then trivially true for $pgt 2$ and easy for $p=2$. But really the statement here is an early and elementary observation in the journey towards proving the last theorem.
$endgroup$
– Mark Bennet
Jan 29 at 7:43
add a comment |
$begingroup$
Fermat's Last Theorem is a distraction here - it says, amongst other things, that there are no solutions to the original for $pgt 2$. A false statement implies anything, so the implication is then trivially true for $pgt 2$ and easy for $p=2$. But really the statement here is an early and elementary observation in the journey towards proving the last theorem.
$endgroup$
– Mark Bennet
Jan 29 at 7:43
$begingroup$
Fermat's Last Theorem is a distraction here - it says, amongst other things, that there are no solutions to the original for $pgt 2$. A false statement implies anything, so the implication is then trivially true for $pgt 2$ and easy for $p=2$. But really the statement here is an early and elementary observation in the journey towards proving the last theorem.
$endgroup$
– Mark Bennet
Jan 29 at 7:43
$begingroup$
Fermat's Last Theorem is a distraction here - it says, amongst other things, that there are no solutions to the original for $pgt 2$. A false statement implies anything, so the implication is then trivially true for $pgt 2$ and easy for $p=2$. But really the statement here is an early and elementary observation in the journey towards proving the last theorem.
$endgroup$
– Mark Bennet
Jan 29 at 7:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Write $$a^p+b^p-c^p=(a^p-a)+(b^p-b)-(c^p-c)+a+b-c.$$
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
By Fermat's little theorem, $0equiv a^p+b^p-c^pequiv a+b-cpmod p$.
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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active
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active
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$begingroup$
Write $$a^p+b^p-c^p=(a^p-a)+(b^p-b)-(c^p-c)+a+b-c.$$
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Write $$a^p+b^p-c^p=(a^p-a)+(b^p-b)-(c^p-c)+a+b-c.$$
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Write $$a^p+b^p-c^p=(a^p-a)+(b^p-b)-(c^p-c)+a+b-c.$$
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Write $$a^p+b^p-c^p=(a^p-a)+(b^p-b)-(c^p-c)+a+b-c.$$
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 6:41
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
$begingroup$
By Fermat's little theorem, $0equiv a^p+b^p-c^pequiv a+b-cpmod p$.
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
By Fermat's little theorem, $0equiv a^p+b^p-c^pequiv a+b-cpmod p$.
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
By Fermat's little theorem, $0equiv a^p+b^p-c^pequiv a+b-cpmod p$.
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
$endgroup$
By Fermat's little theorem, $0equiv a^p+b^p-c^pequiv a+b-cpmod p$.
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
edited Jan 29 at 15:03
J. W. Tanner
4,0611320
4,0611320
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 6:48
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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$begingroup$
Fermat's Last Theorem is a distraction here - it says, amongst other things, that there are no solutions to the original for $pgt 2$. A false statement implies anything, so the implication is then trivially true for $pgt 2$ and easy for $p=2$. But really the statement here is an early and elementary observation in the journey towards proving the last theorem.
$endgroup$
– Mark Bennet
Jan 29 at 7:43