Prove that $1^2 cdot 3^2 cdot 5^2cdots (p-2)^2equiv (-1)^{frac{p+1}{2}} pmod p$
$1^2 cdot 3^2 cdot 5^2cdots (p-2)^2equiv (-1)^{frac{p+1}{2}}pmod p$
I saw that I can use Wilson's theorem that $(p-1)!equiv -1 pmod p$ and that they change something, they said put on even number $frac{p+1}{2}$ so now you prove it but I do not understand can you help me?
discrete-mathematics
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
asked Nov 25 '18 at 20:31
Marko Škorić
70310
add a comment |
$1^2 cdot 3^2 cdot 5^2cdots (p-2)^2equiv (-1)^{frac{p+1}{2}}pmod p$
I saw that I can use Wilson's theorem that $(p-1)!equiv -1 pmod p$ and that they change something, they said put on even number $frac{p+1}{2}$ so now you prove it but I do not understand can you help me?
discrete-mathematics
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
asked Nov 25 '18 at 20:31
Marko Škorić
70310
add a comment |
$1^2 cdot 3^2 cdot 5^2cdots (p-2)^2equiv (-1)^{frac{p+1}{2}}pmod p$
I saw that I can use Wilson's theorem that $(p-1)!equiv -1 pmod p$ and that they change something, they said put on even number $frac{p+1}{2}$ so now you prove it but I do not understand can you help me?
discrete-mathematics
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
asked Nov 25 '18 at 20:31
Marko Škorić
70310
$1^2 cdot 3^2 cdot 5^2cdots (p-2)^2equiv (-1)^{frac{p+1}{2}}pmod p$
I saw that I can use Wilson's theorem that $(p-1)!equiv -1 pmod p$ and that they change something, they said put on even number $frac{p+1}{2}$ so now you prove it but I do not understand can you help me?
discrete-mathematics
discrete-mathematics
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
asked Nov 25 '18 at 20:31
Marko Škorić
70310
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
asked Nov 25 '18 at 20:31
Marko Škorić
70310
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
edited Nov 25 '18 at 20:47
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Nov 25 '18 at 20:31
Marko Škorić
70310
asked Nov 25 '18 at 20:31
Marko Škorić
70310
asked Nov 25 '18 at 20:31
Marko Škorić
70310
70310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
We work in the field $Bbb F_p$ with $p$ elements. (Same as $Bbb Z/p$, for the given prime $p$, i suppose it is a prime.)
Then
$$
begin{aligned}
-1 &= 1cdot 2cdot 3cdot 4cdot 5cdot 6cdot dots cdot (p-2)cdot(p-1)
\
&=1color{red}{cdot (-2)}cdot 3color{red}{cdot (-4)}cdot 5color{red}{cdot (-6)}cdot dots cdot (p-2)color{red}{cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (-2)cdot (-4)cdot (-6)cdot dots cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (p-2)cdot (p-4)cdot (p-6)cdot dots cdot(p-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=(1cdot 3cdot 5cdot dots cdot (p-2))^2
color{blue}{cdot(-1)^{(p-1)/2}}
.
end{aligned}
$$
This leads to the answer...
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013352%2fprove-that-12-cdot-32-cdot-52-cdots-p-22-equiv-1-fracp12-pm%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
We work in the field $Bbb F_p$ with $p$ elements. (Same as $Bbb Z/p$, for the given prime $p$, i suppose it is a prime.)
Then
$$
begin{aligned}
-1 &= 1cdot 2cdot 3cdot 4cdot 5cdot 6cdot dots cdot (p-2)cdot(p-1)
\
&=1color{red}{cdot (-2)}cdot 3color{red}{cdot (-4)}cdot 5color{red}{cdot (-6)}cdot dots cdot (p-2)color{red}{cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (-2)cdot (-4)cdot (-6)cdot dots cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (p-2)cdot (p-4)cdot (p-6)cdot dots cdot(p-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=(1cdot 3cdot 5cdot dots cdot (p-2))^2
color{blue}{cdot(-1)^{(p-1)/2}}
.
end{aligned}
$$
This leads to the answer...
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
add a comment |
We work in the field $Bbb F_p$ with $p$ elements. (Same as $Bbb Z/p$, for the given prime $p$, i suppose it is a prime.)
Then
$$
begin{aligned}
-1 &= 1cdot 2cdot 3cdot 4cdot 5cdot 6cdot dots cdot (p-2)cdot(p-1)
\
&=1color{red}{cdot (-2)}cdot 3color{red}{cdot (-4)}cdot 5color{red}{cdot (-6)}cdot dots cdot (p-2)color{red}{cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (-2)cdot (-4)cdot (-6)cdot dots cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (p-2)cdot (p-4)cdot (p-6)cdot dots cdot(p-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=(1cdot 3cdot 5cdot dots cdot (p-2))^2
color{blue}{cdot(-1)^{(p-1)/2}}
.
end{aligned}
$$
This leads to the answer...
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
add a comment |
We work in the field $Bbb F_p$ with $p$ elements. (Same as $Bbb Z/p$, for the given prime $p$, i suppose it is a prime.)
Then
$$
begin{aligned}
-1 &= 1cdot 2cdot 3cdot 4cdot 5cdot 6cdot dots cdot (p-2)cdot(p-1)
\
&=1color{red}{cdot (-2)}cdot 3color{red}{cdot (-4)}cdot 5color{red}{cdot (-6)}cdot dots cdot (p-2)color{red}{cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (-2)cdot (-4)cdot (-6)cdot dots cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (p-2)cdot (p-4)cdot (p-6)cdot dots cdot(p-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=(1cdot 3cdot 5cdot dots cdot (p-2))^2
color{blue}{cdot(-1)^{(p-1)/2}}
.
end{aligned}
$$
This leads to the answer...
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
We work in the field $Bbb F_p$ with $p$ elements. (Same as $Bbb Z/p$, for the given prime $p$, i suppose it is a prime.)
Then
$$
begin{aligned}
-1 &= 1cdot 2cdot 3cdot 4cdot 5cdot 6cdot dots cdot (p-2)cdot(p-1)
\
&=1color{red}{cdot (-2)}cdot 3color{red}{cdot (-4)}cdot 5color{red}{cdot (-6)}cdot dots cdot (p-2)color{red}{cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (-2)cdot (-4)cdot (-6)cdot dots cdot(-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=1cdot 3cdot 5cdot dots cdot (p-2)
color{red}{cdot (p-2)cdot (p-4)cdot (p-6)cdot dots cdot(p-(p-1))}
color{blue}{cdot(-1)^{(p-1)/2}}
\
&=(1cdot 3cdot 5cdot dots cdot (p-2))^2
color{blue}{cdot(-1)^{(p-1)/2}}
.
end{aligned}
$$
This leads to the answer...
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
answered Nov 25 '18 at 20:44
dan_fulea
6,2301312
6,2301312
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013352%2fprove-that-12-cdot-32-cdot-52-cdots-p-22-equiv-1-fracp12-pm%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
