Question about Wilson's theoremHelp manipulating Wilson's Theorem.Wilson's Theorem and related identitiesWilson's theoremPart of a proof for Wilson's TheoremWilson's theorem intuition(Elementary Number Theory) Using Wilson's Theorem ..??Using Wilson's Theorem to prove Fermat's Little TheoremProve $717$ is not prime using Wilson's TheoremProving Wilson's theoremQuestion about Wilson's theorem, when $n = 4$.
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Question about Wilson's theorem
Help manipulating Wilson's Theorem.Wilson's Theorem and related identitiesWilson's theoremPart of a proof for Wilson's TheoremWilson's theorem intuition(Elementary Number Theory) Using Wilson's Theorem ..??Using Wilson's Theorem to prove Fermat's Little TheoremProve $717$ is not prime using Wilson's TheoremProving Wilson's theoremQuestion about Wilson's theorem, when $n = 4$.
$begingroup$
For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?
Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.
However is $p in mathbbZ$? Or is $p in mathbbN$?
abstract-algebra elementary-number-theory
$endgroup$
add a comment |
$begingroup$
For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?
Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.
However is $p in mathbbZ$? Or is $p in mathbbN$?
abstract-algebra elementary-number-theory
$endgroup$
2
$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08
add a comment |
$begingroup$
For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?
Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.
However is $p in mathbbZ$? Or is $p in mathbbN$?
abstract-algebra elementary-number-theory
$endgroup$
For Wilson's theorem, if $p$ is prime then $(p-1) equiv -1 modp$ and if not $0 modp$ except for $p=4$, is $p$ and integer or natural number?
Studying Wilson's theorem for Double, Hyper, Sub and Double factorials and I have begun by stating what Wilson's theorem actually is.
However is $p in mathbbZ$? Or is $p in mathbbN$?
abstract-algebra elementary-number-theory
abstract-algebra elementary-number-theory
edited Mar 11 at 9:55
Andrews
1,2691421
1,2691421
asked Mar 11 at 9:45
J.BryJ.Bry
486
486
2
$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08
add a comment |
2
$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08
2
2
$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08
$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08
add a comment |
1 Answer
1
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$begingroup$
I assume you mean $(p-1)!$
It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:
$(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).
More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.
New contributor
$endgroup$
add a comment |
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$begingroup$
I assume you mean $(p-1)!$
It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:
$(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).
More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.
New contributor
$endgroup$
add a comment |
$begingroup$
I assume you mean $(p-1)!$
It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:
$(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).
More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.
New contributor
$endgroup$
add a comment |
$begingroup$
I assume you mean $(p-1)!$
It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:
$(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).
More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.
New contributor
$endgroup$
I assume you mean $(p-1)!$
It is not a loss of generality to assume $p in mathbbN$, as primes are unique upto multiplication by units (in this case $pm 1$). The factorial function is not defined at negative integers. However, you may restate the theorem for $-p$ as:
$(-1)(-2)...(-p+1)$ is congruent to $-1$ modulo $-p$ if $-p$ is prime and $0$ otherwise (except $-p = -4$).
More specifically, $p$ and $-p$ generate the same ideal in $mathbbZ$, so the quotient fields are isomorphic.
New contributor
New contributor
answered Mar 11 at 10:06
nammienammie
1899
1899
New contributor
New contributor
add a comment |
add a comment |
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$begingroup$
It's $(p-1)!$ : how do you define factorials for negative integers ?
$endgroup$
– Max
Mar 11 at 10:08