About a $p$-th power residue
$begingroup$
A sentence I was reading reads "Let $p$ be a prime factor of $n$. Take another prime $l$ with $lequiv1pmod{2p}$ so that $-1$ is a $p$-th power residue mod $l$."
Was it saying that for any prime $l$ satisfying $lequiv1pmod{2p}$, $-1$ is a $p$-th power residue mod $l$? Or was it saying to take a prime $l$ such that both (1) $lequiv1pmod{2p}$ and (2) $-1$ is a $p$-th power residue mod $l$ hold?
elementary-number-theory
asked Dec 26 '18 at 13:01
saisai
1376
$endgroup$
add a comment |
$begingroup$
A sentence I was reading reads "Let $p$ be a prime factor of $n$. Take another prime $l$ with $lequiv1pmod{2p}$ so that $-1$ is a $p$-th power residue mod $l$."
Was it saying that for any prime $l$ satisfying $lequiv1pmod{2p}$, $-1$ is a $p$-th power residue mod $l$? Or was it saying to take a prime $l$ such that both (1) $lequiv1pmod{2p}$ and (2) $-1$ is a $p$-th power residue mod $l$ hold?
elementary-number-theory
asked Dec 26 '18 at 13:01
saisai
1376
$endgroup$
$begingroup$
If $p$ is odd then $(-1) = (-1)^p$, so the hypothesis of $-1$ being a $p$th power residue automatically holds in a simple way; it has no nontrivial content.
$endgroup$
– KCd
Dec 26 '18 at 14:08
add a comment |
$begingroup$
A sentence I was reading reads "Let $p$ be a prime factor of $n$. Take another prime $l$ with $lequiv1pmod{2p}$ so that $-1$ is a $p$-th power residue mod $l$."
Was it saying that for any prime $l$ satisfying $lequiv1pmod{2p}$, $-1$ is a $p$-th power residue mod $l$? Or was it saying to take a prime $l$ such that both (1) $lequiv1pmod{2p}$ and (2) $-1$ is a $p$-th power residue mod $l$ hold?
elementary-number-theory
asked Dec 26 '18 at 13:01
saisai
1376
$endgroup$
A sentence I was reading reads "Let $p$ be a prime factor of $n$. Take another prime $l$ with $lequiv1pmod{2p}$ so that $-1$ is a $p$-th power residue mod $l$."
Was it saying that for any prime $l$ satisfying $lequiv1pmod{2p}$, $-1$ is a $p$-th power residue mod $l$? Or was it saying to take a prime $l$ such that both (1) $lequiv1pmod{2p}$ and (2) $-1$ is a $p$-th power residue mod $l$ hold?
elementary-number-theory
elementary-number-theory
asked Dec 26 '18 at 13:01
saisai
1376
asked Dec 26 '18 at 13:01
saisai
1376
asked Dec 26 '18 at 13:01
saisai
1376
asked Dec 26 '18 at 13:01
saisai
1376
asked Dec 26 '18 at 13:01
saisai
1376
1376
$begingroup$
If $p$ is odd then $(-1) = (-1)^p$, so the hypothesis of $-1$ being a $p$th power residue automatically holds in a simple way; it has no nontrivial content.
$endgroup$
– KCd
Dec 26 '18 at 14:08
add a comment |
$begingroup$
If $p$ is odd then $(-1) = (-1)^p$, so the hypothesis of $-1$ being a $p$th power residue automatically holds in a simple way; it has no nontrivial content.
$endgroup$
– KCd
Dec 26 '18 at 14:08
$begingroup$
If $p$ is odd then $(-1) = (-1)^p$, so the hypothesis of $-1$ being a $p$th power residue automatically holds in a simple way; it has no nontrivial content.
$endgroup$
– KCd
Dec 26 '18 at 14:08
$begingroup$
If $p$ is odd then $(-1) = (-1)^p$, so the hypothesis of $-1$ being a $p$th power residue automatically holds in a simple way; it has no nontrivial content.
$endgroup$
– KCd
Dec 26 '18 at 14:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $l$ is odd and $r$ is a primitive root modulo $l$, then
$r^{l-1}equiv 1 pmod{l} $. So if $x = r^{(l-1)/2},$ we have $x^2 equiv 1 pmod{l}.$ The only solutions are $xequiv pm 1 pmod{l}$, but since $r$ is a primitive root, we must have $xequiv -1 pmod{l}.$
Now note that $lequiv 1 pmod{2p}$ means that $l-1 = 2pd$ for some $d,$ which can be written $(l-1)/2 =pd.$ Then you have
$$-1 equiv x equiv r^{(l-1)/2} equiv r^{pd} equiv (r^d)^p pmod{l}, $$
showing that your first condition implies the second.
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
$endgroup$
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
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%2f3052918%2fabout-a-p-th-power-residue%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
$begingroup$
If $l$ is odd and $r$ is a primitive root modulo $l$, then
$r^{l-1}equiv 1 pmod{l} $. So if $x = r^{(l-1)/2},$ we have $x^2 equiv 1 pmod{l}.$ The only solutions are $xequiv pm 1 pmod{l}$, but since $r$ is a primitive root, we must have $xequiv -1 pmod{l}.$
Now note that $lequiv 1 pmod{2p}$ means that $l-1 = 2pd$ for some $d,$ which can be written $(l-1)/2 =pd.$ Then you have
$$-1 equiv x equiv r^{(l-1)/2} equiv r^{pd} equiv (r^d)^p pmod{l}, $$
showing that your first condition implies the second.
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
$endgroup$
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
add a comment |
$begingroup$
If $l$ is odd and $r$ is a primitive root modulo $l$, then
$r^{l-1}equiv 1 pmod{l} $. So if $x = r^{(l-1)/2},$ we have $x^2 equiv 1 pmod{l}.$ The only solutions are $xequiv pm 1 pmod{l}$, but since $r$ is a primitive root, we must have $xequiv -1 pmod{l}.$
Now note that $lequiv 1 pmod{2p}$ means that $l-1 = 2pd$ for some $d,$ which can be written $(l-1)/2 =pd.$ Then you have
$$-1 equiv x equiv r^{(l-1)/2} equiv r^{pd} equiv (r^d)^p pmod{l}, $$
showing that your first condition implies the second.
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
$endgroup$
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
add a comment |
$begingroup$
If $l$ is odd and $r$ is a primitive root modulo $l$, then
$r^{l-1}equiv 1 pmod{l} $. So if $x = r^{(l-1)/2},$ we have $x^2 equiv 1 pmod{l}.$ The only solutions are $xequiv pm 1 pmod{l}$, but since $r$ is a primitive root, we must have $xequiv -1 pmod{l}.$
Now note that $lequiv 1 pmod{2p}$ means that $l-1 = 2pd$ for some $d,$ which can be written $(l-1)/2 =pd.$ Then you have
$$-1 equiv x equiv r^{(l-1)/2} equiv r^{pd} equiv (r^d)^p pmod{l}, $$
showing that your first condition implies the second.
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
$endgroup$
If $l$ is odd and $r$ is a primitive root modulo $l$, then
$r^{l-1}equiv 1 pmod{l} $. So if $x = r^{(l-1)/2},$ we have $x^2 equiv 1 pmod{l}.$ The only solutions are $xequiv pm 1 pmod{l}$, but since $r$ is a primitive root, we must have $xequiv -1 pmod{l}.$
Now note that $lequiv 1 pmod{2p}$ means that $l-1 = 2pd$ for some $d,$ which can be written $(l-1)/2 =pd.$ Then you have
$$-1 equiv x equiv r^{(l-1)/2} equiv r^{pd} equiv (r^d)^p pmod{l}, $$
showing that your first condition implies the second.
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
19.1k21441
19.1k21441
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
add a comment |
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
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.
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%2f3052918%2fabout-a-p-th-power-residue%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
$begingroup$
If $p$ is odd then $(-1) = (-1)^p$, so the hypothesis of $-1$ being a $p$th power residue automatically holds in a simple way; it has no nontrivial content.
$endgroup$
– KCd
Dec 26 '18 at 14:08