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
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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.
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
18.9k21440
$endgroup$
$begingroup$
Thank you very much! This is easy to understand.
$endgroup$
– sai
Dec 27 '18 at 0:29
add a comment |
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1 Answer
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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.
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
18.9k21440
$endgroup$
$begingroup$
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
18.9k21440
$endgroup$
$begingroup$
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
18.9k21440
$endgroup$
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
18.9k21440
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
18.9k21440
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
18.9k21440
answered Dec 26 '18 at 17:47
B. GoddardB. Goddard
18.9k21440
18.9k21440
$begingroup$
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 |
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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