Mixing
Show that $a^{(p-1)!+1} equiv a mod p$
$begingroup$
I'm working in the following excercise:
Show that $a^{(p-1)!+1} equiv a mod p$
By Wilson's theorem I have:
$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$
I'm not sure about my proof, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
asked Jan 30 at 21:57
mrazmraz
450110
$endgroup$
add a comment |
$begingroup$
I'm working in the following excercise:
Show that $a^{(p-1)!+1} equiv a mod p$
By Wilson's theorem I have:
$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$
I'm not sure about my proof, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
asked Jan 30 at 21:57
mrazmraz
450110
$endgroup$
3
$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05
1
$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06
1
$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08
add a comment |
$begingroup$
I'm working in the following excercise:
Show that $a^{(p-1)!+1} equiv a mod p$
By Wilson's theorem I have:
$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$
I'm not sure about my proof, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
asked Jan 30 at 21:57
mrazmraz
450110
$endgroup$
I'm working in the following excercise:
Show that $a^{(p-1)!+1} equiv a mod p$
By Wilson's theorem I have:
$$a^{(p-1)!} equiv -1 mod p$$
$$a^{(p-1)!} * a equiv -1 * amod p$$
$$a^{(p-1)!+1} equiv -a mod p$$
$$a^{(p-1)!+1} equiv a mod p$$
I'm not sure about my proof, is that correct? any help will be really appreciated.
number-theory modular-arithmetic
number-theory modular-arithmetic
asked Jan 30 at 21:57
mrazmraz
450110
asked Jan 30 at 21:57
mrazmraz
450110
asked Jan 30 at 21:57
mrazmraz
450110
asked Jan 30 at 21:57
mrazmraz
450110
asked Jan 30 at 21:57
mrazmraz
450110
450110
3
$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05
1
$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06
1
$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08
add a comment |
3
$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05
1
$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06
1
$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08
3
3
$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05
$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05
1
1
$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06
$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06
1
1
$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08
$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$
[In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]
answered Jan 30 at 22:07
MikeMike
4,611512
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$
[In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]
answered Jan 30 at 22:07
MikeMike
4,611512
$endgroup$
add a comment |
$begingroup$
Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$
[In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]
answered Jan 30 at 22:07
MikeMike
4,611512
$endgroup$
add a comment |
$begingroup$
Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$
[In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]
answered Jan 30 at 22:07
MikeMike
4,611512
$endgroup$
Actually it is $a^{(p-1)!} equiv 1$ mod $p$ (not $-1$). Indeed, $a^{k(p-1)} equiv 1$ for any integer $k$, as $|(mathbb{F}_p)^{times}| = p-1.$
[In general, let $G$ be any group. Then for every element $g in G$, the equation $g^{k|G|} = e$ holds for any positive integer $k$, where $e$ is the identity element. Here set $G = (mathbb{F}_p)^{times}$ where the operation is multiplcation.]
answered Jan 30 at 22:07
MikeMike
4,611512
answered Jan 30 at 22:07
MikeMike
4,611512
answered Jan 30 at 22:07
MikeMike
4,611512
answered Jan 30 at 22:07
MikeMike
4,611512
4,611512
add a comment |
add a comment |
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3
$begingroup$
Why can you suddenly rid of the minus sign?
$endgroup$
– Randall
Jan 30 at 22:05
1
$begingroup$
How can you justify that $-aequiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:06
1
$begingroup$
How does wilson's th which states $(p-1)! equiv -1 pmod p$ make you assume $a^{(p-1)!}equiv a pmod p$?
$endgroup$
– fleablood
Jan 30 at 22:08