show that $(1^p)^k+(2^p)^k+cdots+((p-1)^p)^k$ is divisible by $p$
0
$begingroup$
This will be reduced to $1^k+2^k+3^k+cdots+(p-1)^k$ by Fermat Theorem. I know that the sum of the reduced residue system of $n$ is divisible by $n$. What I need to show, though, is that $1^k+2^k+3^k+cdots+(p-1)^k$ is a reduced residue system of $p$. How to do so?
Is there any other methods to prove so?
Added
I thought of pairing elements such as $a$ and $p-a equiv -a$ to cancelled out. But this works for odd $k$ only.
number-theory elementary-number-theory modular-arithmetic divisibility
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
$endgroup$
|
show 13 more comments
0
$begingroup$
This will be reduced to $1^k+2^k+3^k+cdots+(p-1)^k$ by Fermat Theorem. I know that the sum of the reduced residue system of $n$ is divisible by $n$. What I need to show, though, is that $1^k+2^k+3^k+cdots+(p-1)^k$ is a reduced residue system of $p$. How to do so?
Is there any other methods to prove so?
Added
I thought of pairing elements such as $a$ and $p-a equiv -a$ to cancelled out. But this works for odd $k$ only.
number-theory elementary-number-theory modular-arithmetic divisibility
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
$endgroup$
2
$begingroup$
This is not clear. Are you claiming that, say, $2^{p^k}equiv 2^kpmod p$? But this is false...$2^{3^2}=2^9equiv 2not equiv 2^2pmod 3$.
$endgroup$
– lulu
Dec 20 '18 at 16:54
4
$begingroup$
No. Keep in mind that $a^{p^k}$ denotes $a^{(p^k)}$ and not $left( a^pright)^k$ (which, after all, would just be $a^{pk}$).
$endgroup$
– lulu
Dec 20 '18 at 17:01
2
$begingroup$
This problem is actually easier if you don't reduce using Fermat's theorem. You were on the right path with pairing, and for all primes greater than 2 your exponents will be odd. The case of 2 is a trivial case though.
$endgroup$
– mascoj
Dec 20 '18 at 17:01
1
$begingroup$
It is not a reduced residue system for $p=7$ and $k=3$.
$endgroup$
– lhf
Dec 20 '18 at 17:07
6
$begingroup$
Possible duplicate of $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$?
$endgroup$
– lulu
Dec 20 '18 at 17:17
|
show 13 more comments
0
0
0
$begingroup$
This will be reduced to $1^k+2^k+3^k+cdots+(p-1)^k$ by Fermat Theorem. I know that the sum of the reduced residue system of $n$ is divisible by $n$. What I need to show, though, is that $1^k+2^k+3^k+cdots+(p-1)^k$ is a reduced residue system of $p$. How to do so?
Is there any other methods to prove so?
Added
I thought of pairing elements such as $a$ and $p-a equiv -a$ to cancelled out. But this works for odd $k$ only.
number-theory elementary-number-theory modular-arithmetic divisibility
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
$endgroup$
This will be reduced to $1^k+2^k+3^k+cdots+(p-1)^k$ by Fermat Theorem. I know that the sum of the reduced residue system of $n$ is divisible by $n$. What I need to show, though, is that $1^k+2^k+3^k+cdots+(p-1)^k$ is a reduced residue system of $p$. How to do so?
Is there any other methods to prove so?
Added
I thought of pairing elements such as $a$ and $p-a equiv -a$ to cancelled out. But this works for odd $k$ only.
number-theory elementary-number-theory modular-arithmetic divisibility
number-theory elementary-number-theory modular-arithmetic divisibility
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
edited Dec 20 '18 at 17:02
Maged Saeed
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
asked Dec 20 '18 at 16:48
Maged SaeedMaged Saeed
8771417
8771417
2
$begingroup$
This is not clear. Are you claiming that, say, $2^{p^k}equiv 2^kpmod p$? But this is false...$2^{3^2}=2^9equiv 2not equiv 2^2pmod 3$.
$endgroup$
– lulu
Dec 20 '18 at 16:54
4
$begingroup$
No. Keep in mind that $a^{p^k}$ denotes $a^{(p^k)}$ and not $left( a^pright)^k$ (which, after all, would just be $a^{pk}$).
$endgroup$
– lulu
Dec 20 '18 at 17:01
2
$begingroup$
This problem is actually easier if you don't reduce using Fermat's theorem. You were on the right path with pairing, and for all primes greater than 2 your exponents will be odd. The case of 2 is a trivial case though.
$endgroup$
– mascoj
Dec 20 '18 at 17:01
1
$begingroup$
It is not a reduced residue system for $p=7$ and $k=3$.
$endgroup$
– lhf
Dec 20 '18 at 17:07
6
$begingroup$
Possible duplicate of $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$?
$endgroup$
– lulu
Dec 20 '18 at 17:17
|
show 13 more comments
2
$begingroup$
This is not clear. Are you claiming that, say, $2^{p^k}equiv 2^kpmod p$? But this is false...$2^{3^2}=2^9equiv 2not equiv 2^2pmod 3$.
$endgroup$
– lulu
Dec 20 '18 at 16:54
4
$begingroup$
No. Keep in mind that $a^{p^k}$ denotes $a^{(p^k)}$ and not $left( a^pright)^k$ (which, after all, would just be $a^{pk}$).
$endgroup$
– lulu
Dec 20 '18 at 17:01
2
$begingroup$
This problem is actually easier if you don't reduce using Fermat's theorem. You were on the right path with pairing, and for all primes greater than 2 your exponents will be odd. The case of 2 is a trivial case though.
$endgroup$
– mascoj
Dec 20 '18 at 17:01
1
$begingroup$
It is not a reduced residue system for $p=7$ and $k=3$.
$endgroup$
– lhf
Dec 20 '18 at 17:07
6
$begingroup$
Possible duplicate of $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$?
$endgroup$
– lulu
Dec 20 '18 at 17:17
2
2
$begingroup$
This is not clear. Are you claiming that, say, $2^{p^k}equiv 2^kpmod p$? But this is false...$2^{3^2}=2^9equiv 2not equiv 2^2pmod 3$.
$endgroup$
– lulu
Dec 20 '18 at 16:54
$begingroup$
This is not clear. Are you claiming that, say, $2^{p^k}equiv 2^kpmod p$? But this is false...$2^{3^2}=2^9equiv 2not equiv 2^2pmod 3$.
$endgroup$
– lulu
Dec 20 '18 at 16:54
4
4
$begingroup$
No. Keep in mind that $a^{p^k}$ denotes $a^{(p^k)}$ and not $left( a^pright)^k$ (which, after all, would just be $a^{pk}$).
$endgroup$
– lulu
Dec 20 '18 at 17:01
$begingroup$
No. Keep in mind that $a^{p^k}$ denotes $a^{(p^k)}$ and not $left( a^pright)^k$ (which, after all, would just be $a^{pk}$).
$endgroup$
– lulu
Dec 20 '18 at 17:01
2
2
$begingroup$
This problem is actually easier if you don't reduce using Fermat's theorem. You were on the right path with pairing, and for all primes greater than 2 your exponents will be odd. The case of 2 is a trivial case though.
$endgroup$
– mascoj
Dec 20 '18 at 17:01
$begingroup$
This problem is actually easier if you don't reduce using Fermat's theorem. You were on the right path with pairing, and for all primes greater than 2 your exponents will be odd. The case of 2 is a trivial case though.
$endgroup$
– mascoj
Dec 20 '18 at 17:01
1
1
$begingroup$
It is not a reduced residue system for $p=7$ and $k=3$.
$endgroup$
– lhf
Dec 20 '18 at 17:07
$begingroup$
It is not a reduced residue system for $p=7$ and $k=3$.
$endgroup$
– lhf
Dec 20 '18 at 17:07
6
6
$begingroup$
Possible duplicate of $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$?
$endgroup$
– lulu
Dec 20 '18 at 17:17
$begingroup$
Possible duplicate of $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$?
$endgroup$
– lulu
Dec 20 '18 at 17:17
|
show 13 more comments
0
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2
$begingroup$
This is not clear. Are you claiming that, say, $2^{p^k}equiv 2^kpmod p$? But this is false...$2^{3^2}=2^9equiv 2not equiv 2^2pmod 3$.
$endgroup$
– lulu
Dec 20 '18 at 16:54
4
$begingroup$
No. Keep in mind that $a^{p^k}$ denotes $a^{(p^k)}$ and not $left( a^pright)^k$ (which, after all, would just be $a^{pk}$).
$endgroup$
– lulu
Dec 20 '18 at 17:01
2
$begingroup$
This problem is actually easier if you don't reduce using Fermat's theorem. You were on the right path with pairing, and for all primes greater than 2 your exponents will be odd. The case of 2 is a trivial case though.
$endgroup$
– mascoj
Dec 20 '18 at 17:01
1
$begingroup$
It is not a reduced residue system for $p=7$ and $k=3$.
$endgroup$
– lhf
Dec 20 '18 at 17:07
6
$begingroup$
Possible duplicate of $ 1^k+2^k+3^k+...+(p-1)^k $ always a multiple of $p$?
$endgroup$
– lulu
Dec 20 '18 at 17:17