About a $p$-th power residue












0












$begingroup$


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







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  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0


0



$begingroup$


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







share|cite|improve this question









$endgroup$




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






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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


















  • $begingroup$
    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










1 Answer
1






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$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much! This is easy to understand.
    $endgroup$
    – sai
    Dec 27 '18 at 0:29











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1 Answer
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1 Answer
1






active

oldest

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active

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active

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0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much! This is easy to understand.
    $endgroup$
    – sai
    Dec 27 '18 at 0:29
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much! This is easy to understand.
    $endgroup$
    – sai
    Dec 27 '18 at 0:29














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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