Legendre symbol of $-n$
0
$begingroup$
Let $x, y, n, p$ be positive integers such that
$$x^2 + ny^2 = p,$$
where $p$ is a prime such that $p neq n$. I need to show that the Legendre symbol $left(frac{-n}{p}right)=1$.
elementary-number-theory prime-numbers
edited Dec 17 '18 at 16:57
Klangen
1,71011334
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
$endgroup$
add a comment |
0
$begingroup$
Let $x, y, n, p$ be positive integers such that
$$x^2 + ny^2 = p,$$
where $p$ is a prime such that $p neq n$. I need to show that the Legendre symbol $left(frac{-n}{p}right)=1$.
elementary-number-theory prime-numbers
edited Dec 17 '18 at 16:57
Klangen
1,71011334
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
$endgroup$
$begingroup$
neither $x$ nor $y$ is divisible by $p,$ if only because that would make $x^2 + n y^2$ strictly larger than $p.$
$endgroup$
– Will Jagy
Dec 15 '15 at 23:30
3
$begingroup$
I now get that $x^2 = -ny^2 (mod p)$. Since $y$ and $p$ are relatively prime, we can multiply by the inverse of y and hence $-n = (xy')^2 (mod p)$ and hence is $-n$ a quadratic residue.
$endgroup$
– Ludwwwig
Dec 15 '15 at 23:33
add a comment |
0
0
0
$begingroup$
Let $x, y, n, p$ be positive integers such that
$$x^2 + ny^2 = p,$$
where $p$ is a prime such that $p neq n$. I need to show that the Legendre symbol $left(frac{-n}{p}right)=1$.
elementary-number-theory prime-numbers
edited Dec 17 '18 at 16:57
Klangen
1,71011334
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
$endgroup$
Let $x, y, n, p$ be positive integers such that
$$x^2 + ny^2 = p,$$
where $p$ is a prime such that $p neq n$. I need to show that the Legendre symbol $left(frac{-n}{p}right)=1$.
elementary-number-theory prime-numbers
elementary-number-theory prime-numbers
edited Dec 17 '18 at 16:57
Klangen
1,71011334
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
edited Dec 17 '18 at 16:57
Klangen
1,71011334
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
edited Dec 17 '18 at 16:57
Klangen
1,71011334
edited Dec 17 '18 at 16:57
Klangen
1,71011334
edited Dec 17 '18 at 16:57
Klangen
1,71011334
1,71011334
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
asked Dec 15 '15 at 23:27
LudwwwigLudwwwig
1236
1236
$begingroup$
neither $x$ nor $y$ is divisible by $p,$ if only because that would make $x^2 + n y^2$ strictly larger than $p.$
$endgroup$
– Will Jagy
Dec 15 '15 at 23:30
3
$begingroup$
I now get that $x^2 = -ny^2 (mod p)$. Since $y$ and $p$ are relatively prime, we can multiply by the inverse of y and hence $-n = (xy')^2 (mod p)$ and hence is $-n$ a quadratic residue.
$endgroup$
– Ludwwwig
Dec 15 '15 at 23:33
add a comment |
$begingroup$
neither $x$ nor $y$ is divisible by $p,$ if only because that would make $x^2 + n y^2$ strictly larger than $p.$
$endgroup$
– Will Jagy
Dec 15 '15 at 23:30
3
$begingroup$
I now get that $x^2 = -ny^2 (mod p)$. Since $y$ and $p$ are relatively prime, we can multiply by the inverse of y and hence $-n = (xy')^2 (mod p)$ and hence is $-n$ a quadratic residue.
$endgroup$
– Ludwwwig
Dec 15 '15 at 23:33
$begingroup$
neither $x$ nor $y$ is divisible by $p,$ if only because that would make $x^2 + n y^2$ strictly larger than $p.$
$endgroup$
– Will Jagy
Dec 15 '15 at 23:30
$begingroup$
neither $x$ nor $y$ is divisible by $p,$ if only because that would make $x^2 + n y^2$ strictly larger than $p.$
$endgroup$
– Will Jagy
Dec 15 '15 at 23:30
3
3
$begingroup$
I now get that $x^2 = -ny^2 (mod p)$. Since $y$ and $p$ are relatively prime, we can multiply by the inverse of y and hence $-n = (xy')^2 (mod p)$ and hence is $-n$ a quadratic residue.
$endgroup$
– Ludwwwig
Dec 15 '15 at 23:33
$begingroup$
I now get that $x^2 = -ny^2 (mod p)$. Since $y$ and $p$ are relatively prime, we can multiply by the inverse of y and hence $-n = (xy')^2 (mod p)$ and hence is $-n$ a quadratic residue.
$endgroup$
– Ludwwwig
Dec 15 '15 at 23:33
add a comment |
0
active
oldest
votes
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
});
}
});
draft saved
draft discarded
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%2f1577603%2flegendre-symbol-of-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f1577603%2flegendre-symbol-of-n%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
$begingroup$
neither $x$ nor $y$ is divisible by $p,$ if only because that would make $x^2 + n y^2$ strictly larger than $p.$
$endgroup$
– Will Jagy
Dec 15 '15 at 23:30
3
$begingroup$
I now get that $x^2 = -ny^2 (mod p)$. Since $y$ and $p$ are relatively prime, we can multiply by the inverse of y and hence $-n = (xy')^2 (mod p)$ and hence is $-n$ a quadratic residue.
$endgroup$
– Ludwwwig
Dec 15 '15 at 23:33