When g and -g are both primitive roots
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The question states:
Let $g$ by a primitive root of the odd prime $p$. Show that $-g$ is a primitive root , or not, according as $p equiv 1 pmod 4$ or not.
For me, I cannot see any connection between the type of primes and the primitive root. Any Hint is highly appreciated.
Thanks
number-theory elementary-number-theory modular-arithmetic divisibility primitive-roots
asked Nov 23 at 20:58
Maged Saeed
777316
add a comment |
up vote
1
down vote
favorite
The question states:
Let $g$ by a primitive root of the odd prime $p$. Show that $-g$ is a primitive root , or not, according as $p equiv 1 pmod 4$ or not.
For me, I cannot see any connection between the type of primes and the primitive root. Any Hint is highly appreciated.
Thanks
number-theory elementary-number-theory modular-arithmetic divisibility primitive-roots
asked Nov 23 at 20:58
Maged Saeed
777316
2
Possible duplicate of math.stackexchange.com/questions/1229270/…
– lhf
Nov 25 at 13:06
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The question states:
Let $g$ by a primitive root of the odd prime $p$. Show that $-g$ is a primitive root , or not, according as $p equiv 1 pmod 4$ or not.
For me, I cannot see any connection between the type of primes and the primitive root. Any Hint is highly appreciated.
Thanks
number-theory elementary-number-theory modular-arithmetic divisibility primitive-roots
asked Nov 23 at 20:58
Maged Saeed
777316
The question states:
Let $g$ by a primitive root of the odd prime $p$. Show that $-g$ is a primitive root , or not, according as $p equiv 1 pmod 4$ or not.
For me, I cannot see any connection between the type of primes and the primitive root. Any Hint is highly appreciated.
Thanks
number-theory elementary-number-theory modular-arithmetic divisibility primitive-roots
number-theory elementary-number-theory modular-arithmetic divisibility primitive-roots
asked Nov 23 at 20:58
Maged Saeed
777316
asked Nov 23 at 20:58
Maged Saeed
777316
asked Nov 23 at 20:58
Maged Saeed
777316
asked Nov 23 at 20:58
Maged Saeed
777316
asked Nov 23 at 20:58
Maged Saeed
777316
777316
2
Possible duplicate of math.stackexchange.com/questions/1229270/…
– lhf
Nov 25 at 13:06
add a comment |
2
Possible duplicate of math.stackexchange.com/questions/1229270/…
– lhf
Nov 25 at 13:06
2
2
Possible duplicate of math.stackexchange.com/questions/1229270/…
– lhf
Nov 25 at 13:06
Possible duplicate of math.stackexchange.com/questions/1229270/…
– lhf
Nov 25 at 13:06
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Hint: $-g$ is a primitive root iff $-g = g^k$ with $gcd(k,p-1)=1$. Connect this with the key fact:
$-1$ is a square mod $p$ iff $p equiv 1 bmod 4$
Partial solution:
If $-g$ is a primitive root, then $-g equiv g^k$ with $gcd(k,p-1)=1$ and so $-1 equiv g^{k-1}$. Now $k$ is odd because $gcd(k,p-1)=1$. Therefore, $k-1$ is even and $-1$ is a square mod $p$. Write $-1 equiv a^2$. Then $a$ has order $4$ mod $p$ and so $4$ divides $p-1$, that is $p equiv 1 bmod 4$.
answered Nov 23 at 22:27
lhf
162k9166385
Another hint: is $u$ and $v$ are squares, so is $uv$.
– JavaMan
Nov 23 at 22:48
It seems that I am way beyond understanding even these hints. I will make a review to the topic then come back. Please be available next time :).
– Maged Saeed
Nov 24 at 20:51
Hi @lhf. How $a$ has an order of 4?
– Maged Saeed
Dec 5 at 16:32
1
@MagedSaeed, $a ne1, a^2=-1ne1, a^4=1$.
– lhf
Dec 5 at 17:01
Hi @lhf. Please, consider reading my proof in the answer I have posted to this question. Many Thanks.
– Maged Saeed
Dec 7 at 16:18
add a comment |
up vote
0
down vote
I have came up with this proof after a discussion with a friend of mine. The proof does not use anything from quadratic residue.
First of all, consider this fact:
If $a$ is of order $h$ $pmod n$, then $a^k$ is of order $frac{h}{gcd(h,k)} quad quad quadquad quad (1)$
The Proof:
Since $g$ is a primitive root, $-1 equiv g^{frac{p-1}{2}} pmod p$. Therefore, $-g equiv (-1)(g) equiv g^{frac{p-1}{2}}g equiv g^{frac{p+1}{2}} pmod p$. Now, the order of $g^{frac{p+1}{2}} pmod p$ according to $(1)$ is $frac{p-1}{gcd(frac{p+1}{2},p-1)}$. If $pequiv 1 pmod 4$, then $frac{p+1}{2}$ is odd and ${gcd(frac{p+1}{2},p-1)}$ is 1 making the order of $-g$ to be $p-1$. i.e. a primitive root. Otherwise, the term $frac{p+1}{2}$ is even and ${gcd(frac{p+1}{2},p-1)} > 1$. Therefore, the order of $-g$ is not $p-1$. i.e. not a primitive root.
answered Dec 7 at 16:17
Maged Saeed
777316
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: $-g$ is a primitive root iff $-g = g^k$ with $gcd(k,p-1)=1$. Connect this with the key fact:
$-1$ is a square mod $p$ iff $p equiv 1 bmod 4$
Partial solution:
If $-g$ is a primitive root, then $-g equiv g^k$ with $gcd(k,p-1)=1$ and so $-1 equiv g^{k-1}$. Now $k$ is odd because $gcd(k,p-1)=1$. Therefore, $k-1$ is even and $-1$ is a square mod $p$. Write $-1 equiv a^2$. Then $a$ has order $4$ mod $p$ and so $4$ divides $p-1$, that is $p equiv 1 bmod 4$.
answered Nov 23 at 22:27
lhf
162k9166385
Another hint: is $u$ and $v$ are squares, so is $uv$.
– JavaMan
Nov 23 at 22:48
It seems that I am way beyond understanding even these hints. I will make a review to the topic then come back. Please be available next time :).
– Maged Saeed
Nov 24 at 20:51
Hi @lhf. How $a$ has an order of 4?
– Maged Saeed
Dec 5 at 16:32
1
@MagedSaeed, $a ne1, a^2=-1ne1, a^4=1$.
– lhf
Dec 5 at 17:01
Hi @lhf. Please, consider reading my proof in the answer I have posted to this question. Many Thanks.
– Maged Saeed
Dec 7 at 16:18
add a comment |
up vote
2
down vote
accepted
Hint: $-g$ is a primitive root iff $-g = g^k$ with $gcd(k,p-1)=1$. Connect this with the key fact:
$-1$ is a square mod $p$ iff $p equiv 1 bmod 4$
Partial solution:
If $-g$ is a primitive root, then $-g equiv g^k$ with $gcd(k,p-1)=1$ and so $-1 equiv g^{k-1}$. Now $k$ is odd because $gcd(k,p-1)=1$. Therefore, $k-1$ is even and $-1$ is a square mod $p$. Write $-1 equiv a^2$. Then $a$ has order $4$ mod $p$ and so $4$ divides $p-1$, that is $p equiv 1 bmod 4$.
answered Nov 23 at 22:27
lhf
162k9166385
Another hint: is $u$ and $v$ are squares, so is $uv$.
– JavaMan
Nov 23 at 22:48
It seems that I am way beyond understanding even these hints. I will make a review to the topic then come back. Please be available next time :).
– Maged Saeed
Nov 24 at 20:51
Hi @lhf. How $a$ has an order of 4?
– Maged Saeed
Dec 5 at 16:32
1
@MagedSaeed, $a ne1, a^2=-1ne1, a^4=1$.
– lhf
Dec 5 at 17:01
Hi @lhf. Please, consider reading my proof in the answer I have posted to this question. Many Thanks.
– Maged Saeed
Dec 7 at 16:18
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: $-g$ is a primitive root iff $-g = g^k$ with $gcd(k,p-1)=1$. Connect this with the key fact:
$-1$ is a square mod $p$ iff $p equiv 1 bmod 4$
Partial solution:
If $-g$ is a primitive root, then $-g equiv g^k$ with $gcd(k,p-1)=1$ and so $-1 equiv g^{k-1}$. Now $k$ is odd because $gcd(k,p-1)=1$. Therefore, $k-1$ is even and $-1$ is a square mod $p$. Write $-1 equiv a^2$. Then $a$ has order $4$ mod $p$ and so $4$ divides $p-1$, that is $p equiv 1 bmod 4$.
answered Nov 23 at 22:27
lhf
162k9166385
Hint: $-g$ is a primitive root iff $-g = g^k$ with $gcd(k,p-1)=1$. Connect this with the key fact:
$-1$ is a square mod $p$ iff $p equiv 1 bmod 4$
Partial solution:
If $-g$ is a primitive root, then $-g equiv g^k$ with $gcd(k,p-1)=1$ and so $-1 equiv g^{k-1}$. Now $k$ is odd because $gcd(k,p-1)=1$. Therefore, $k-1$ is even and $-1$ is a square mod $p$. Write $-1 equiv a^2$. Then $a$ has order $4$ mod $p$ and so $4$ divides $p-1$, that is $p equiv 1 bmod 4$.
answered Nov 23 at 22:27
lhf
162k9166385
edited Nov 25 at 13:04
answered Nov 23 at 22:27
lhf
162k9166385
answered Nov 23 at 22:27
lhf
162k9166385
answered Nov 23 at 22:27
lhf
162k9166385
162k9166385
Another hint: is $u$ and $v$ are squares, so is $uv$.
– JavaMan
Nov 23 at 22:48
It seems that I am way beyond understanding even these hints. I will make a review to the topic then come back. Please be available next time :).
– Maged Saeed
Nov 24 at 20:51
Hi @lhf. How $a$ has an order of 4?
– Maged Saeed
Dec 5 at 16:32
1
@MagedSaeed, $a ne1, a^2=-1ne1, a^4=1$.
– lhf
Dec 5 at 17:01
Hi @lhf. Please, consider reading my proof in the answer I have posted to this question. Many Thanks.
– Maged Saeed
Dec 7 at 16:18
add a comment |
Another hint: is $u$ and $v$ are squares, so is $uv$.
– JavaMan
Nov 23 at 22:48
It seems that I am way beyond understanding even these hints. I will make a review to the topic then come back. Please be available next time :).
– Maged Saeed
Nov 24 at 20:51
Hi @lhf. How $a$ has an order of 4?
– Maged Saeed
Dec 5 at 16:32
1
@MagedSaeed, $a ne1, a^2=-1ne1, a^4=1$.
– lhf
Dec 5 at 17:01
Hi @lhf. Please, consider reading my proof in the answer I have posted to this question. Many Thanks.
– Maged Saeed
Dec 7 at 16:18
Another hint: is $u$ and $v$ are squares, so is $uv$.
– JavaMan
Nov 23 at 22:48
Another hint: is $u$ and $v$ are squares, so is $uv$.
– JavaMan
Nov 23 at 22:48
It seems that I am way beyond understanding even these hints. I will make a review to the topic then come back. Please be available next time :).
– Maged Saeed
Nov 24 at 20:51
It seems that I am way beyond understanding even these hints. I will make a review to the topic then come back. Please be available next time :).
– Maged Saeed
Nov 24 at 20:51
Hi @lhf. How $a$ has an order of 4?
– Maged Saeed
Dec 5 at 16:32
Hi @lhf. How $a$ has an order of 4?
– Maged Saeed
Dec 5 at 16:32
1
1
@MagedSaeed, $a ne1, a^2=-1ne1, a^4=1$.
– lhf
Dec 5 at 17:01
@MagedSaeed, $a ne1, a^2=-1ne1, a^4=1$.
– lhf
Dec 5 at 17:01
Hi @lhf. Please, consider reading my proof in the answer I have posted to this question. Many Thanks.
– Maged Saeed
Dec 7 at 16:18
Hi @lhf. Please, consider reading my proof in the answer I have posted to this question. Many Thanks.
– Maged Saeed
Dec 7 at 16:18
add a comment |
up vote
0
down vote
I have came up with this proof after a discussion with a friend of mine. The proof does not use anything from quadratic residue.
First of all, consider this fact:
If $a$ is of order $h$ $pmod n$, then $a^k$ is of order $frac{h}{gcd(h,k)} quad quad quadquad quad (1)$
The Proof:
Since $g$ is a primitive root, $-1 equiv g^{frac{p-1}{2}} pmod p$. Therefore, $-g equiv (-1)(g) equiv g^{frac{p-1}{2}}g equiv g^{frac{p+1}{2}} pmod p$. Now, the order of $g^{frac{p+1}{2}} pmod p$ according to $(1)$ is $frac{p-1}{gcd(frac{p+1}{2},p-1)}$. If $pequiv 1 pmod 4$, then $frac{p+1}{2}$ is odd and ${gcd(frac{p+1}{2},p-1)}$ is 1 making the order of $-g$ to be $p-1$. i.e. a primitive root. Otherwise, the term $frac{p+1}{2}$ is even and ${gcd(frac{p+1}{2},p-1)} > 1$. Therefore, the order of $-g$ is not $p-1$. i.e. not a primitive root.
answered Dec 7 at 16:17
Maged Saeed
777316
add a comment |
up vote
0
down vote
I have came up with this proof after a discussion with a friend of mine. The proof does not use anything from quadratic residue.
First of all, consider this fact:
If $a$ is of order $h$ $pmod n$, then $a^k$ is of order $frac{h}{gcd(h,k)} quad quad quadquad quad (1)$
The Proof:
Since $g$ is a primitive root, $-1 equiv g^{frac{p-1}{2}} pmod p$. Therefore, $-g equiv (-1)(g) equiv g^{frac{p-1}{2}}g equiv g^{frac{p+1}{2}} pmod p$. Now, the order of $g^{frac{p+1}{2}} pmod p$ according to $(1)$ is $frac{p-1}{gcd(frac{p+1}{2},p-1)}$. If $pequiv 1 pmod 4$, then $frac{p+1}{2}$ is odd and ${gcd(frac{p+1}{2},p-1)}$ is 1 making the order of $-g$ to be $p-1$. i.e. a primitive root. Otherwise, the term $frac{p+1}{2}$ is even and ${gcd(frac{p+1}{2},p-1)} > 1$. Therefore, the order of $-g$ is not $p-1$. i.e. not a primitive root.
answered Dec 7 at 16:17
Maged Saeed
777316
add a comment |
up vote
0
down vote
up vote
0
down vote
I have came up with this proof after a discussion with a friend of mine. The proof does not use anything from quadratic residue.
First of all, consider this fact:
If $a$ is of order $h$ $pmod n$, then $a^k$ is of order $frac{h}{gcd(h,k)} quad quad quadquad quad (1)$
The Proof:
Since $g$ is a primitive root, $-1 equiv g^{frac{p-1}{2}} pmod p$. Therefore, $-g equiv (-1)(g) equiv g^{frac{p-1}{2}}g equiv g^{frac{p+1}{2}} pmod p$. Now, the order of $g^{frac{p+1}{2}} pmod p$ according to $(1)$ is $frac{p-1}{gcd(frac{p+1}{2},p-1)}$. If $pequiv 1 pmod 4$, then $frac{p+1}{2}$ is odd and ${gcd(frac{p+1}{2},p-1)}$ is 1 making the order of $-g$ to be $p-1$. i.e. a primitive root. Otherwise, the term $frac{p+1}{2}$ is even and ${gcd(frac{p+1}{2},p-1)} > 1$. Therefore, the order of $-g$ is not $p-1$. i.e. not a primitive root.
answered Dec 7 at 16:17
Maged Saeed
777316
I have came up with this proof after a discussion with a friend of mine. The proof does not use anything from quadratic residue.
First of all, consider this fact:
If $a$ is of order $h$ $pmod n$, then $a^k$ is of order $frac{h}{gcd(h,k)} quad quad quadquad quad (1)$
The Proof:
Since $g$ is a primitive root, $-1 equiv g^{frac{p-1}{2}} pmod p$. Therefore, $-g equiv (-1)(g) equiv g^{frac{p-1}{2}}g equiv g^{frac{p+1}{2}} pmod p$. Now, the order of $g^{frac{p+1}{2}} pmod p$ according to $(1)$ is $frac{p-1}{gcd(frac{p+1}{2},p-1)}$. If $pequiv 1 pmod 4$, then $frac{p+1}{2}$ is odd and ${gcd(frac{p+1}{2},p-1)}$ is 1 making the order of $-g$ to be $p-1$. i.e. a primitive root. Otherwise, the term $frac{p+1}{2}$ is even and ${gcd(frac{p+1}{2},p-1)} > 1$. Therefore, the order of $-g$ is not $p-1$. i.e. not a primitive root.
answered Dec 7 at 16:17
Maged Saeed
777316
edited Dec 7 at 16:35
answered Dec 7 at 16:17
Maged Saeed
777316
answered Dec 7 at 16:17
Maged Saeed
777316
answered Dec 7 at 16:17
Maged Saeed
777316
777316
add a comment |
add a comment |
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2
Possible duplicate of math.stackexchange.com/questions/1229270/…
– lhf
Nov 25 at 13:06