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Classify the odd primes $q$ such that a NEGATIVE number is a quadratic residue $mod{q}$
Suppose we are given $y < -1$. I wish to classify all primes $q$ such that $y$ is a quadratic residue $pmod{q}$, i.e. such that there exists a number $x$ satisfying $$y equiv x^2 pmod{q}.$$
How would I go about doing this? I'm aware that if instead $y>1$ then I can write $y$ as a product of primes and use the law of quadratic reciprocity. However, this doesn't seem to be applicable if $y<-1$, as the Legendre/Jacobi symbol requires a positive number in the denominator, and since we do not know $q$, it seems (although I may be wrong) we cannot find a meaningful positive representative modulo $q$.
elementary-number-theory prime-numbers
asked May 13 '13 at 1:10
Damien
153
add a comment |
Suppose we are given $y < -1$. I wish to classify all primes $q$ such that $y$ is a quadratic residue $pmod{q}$, i.e. such that there exists a number $x$ satisfying $$y equiv x^2 pmod{q}.$$
How would I go about doing this? I'm aware that if instead $y>1$ then I can write $y$ as a product of primes and use the law of quadratic reciprocity. However, this doesn't seem to be applicable if $y<-1$, as the Legendre/Jacobi symbol requires a positive number in the denominator, and since we do not know $q$, it seems (although I may be wrong) we cannot find a meaningful positive representative modulo $q$.
elementary-number-theory prime-numbers
asked May 13 '13 at 1:10
Damien
153
2
You can add $q$ to $y$ as many times as needed to make it positive. Also, you can calculate the Legendre symbol value of $frac{-1}{q}$.
– Calvin Lin
May 13 '13 at 1:14
$left( frac{-1}{q} right)=(-1)^frac{q-1}{2}$ for odd $q$... You would need to study separately what happens when $q$ is even and prime ;)
– N. S.
May 13 '13 at 1:15
add a comment |
Suppose we are given $y < -1$. I wish to classify all primes $q$ such that $y$ is a quadratic residue $pmod{q}$, i.e. such that there exists a number $x$ satisfying $$y equiv x^2 pmod{q}.$$
How would I go about doing this? I'm aware that if instead $y>1$ then I can write $y$ as a product of primes and use the law of quadratic reciprocity. However, this doesn't seem to be applicable if $y<-1$, as the Legendre/Jacobi symbol requires a positive number in the denominator, and since we do not know $q$, it seems (although I may be wrong) we cannot find a meaningful positive representative modulo $q$.
elementary-number-theory prime-numbers
asked May 13 '13 at 1:10
Damien
153
Suppose we are given $y < -1$. I wish to classify all primes $q$ such that $y$ is a quadratic residue $pmod{q}$, i.e. such that there exists a number $x$ satisfying $$y equiv x^2 pmod{q}.$$
How would I go about doing this? I'm aware that if instead $y>1$ then I can write $y$ as a product of primes and use the law of quadratic reciprocity. However, this doesn't seem to be applicable if $y<-1$, as the Legendre/Jacobi symbol requires a positive number in the denominator, and since we do not know $q$, it seems (although I may be wrong) we cannot find a meaningful positive representative modulo $q$.
elementary-number-theory prime-numbers
elementary-number-theory prime-numbers
asked May 13 '13 at 1:10
Damien
153
asked May 13 '13 at 1:10
Damien
153
asked May 13 '13 at 1:10
Damien
153
asked May 13 '13 at 1:10
Damien
153
asked May 13 '13 at 1:10
Damien
153
153
2
You can add $q$ to $y$ as many times as needed to make it positive. Also, you can calculate the Legendre symbol value of $frac{-1}{q}$.
– Calvin Lin
May 13 '13 at 1:14
$left( frac{-1}{q} right)=(-1)^frac{q-1}{2}$ for odd $q$... You would need to study separately what happens when $q$ is even and prime ;)
– N. S.
May 13 '13 at 1:15
add a comment |
2
You can add $q$ to $y$ as many times as needed to make it positive. Also, you can calculate the Legendre symbol value of $frac{-1}{q}$.
– Calvin Lin
May 13 '13 at 1:14
$left( frac{-1}{q} right)=(-1)^frac{q-1}{2}$ for odd $q$... You would need to study separately what happens when $q$ is even and prime ;)
– N. S.
May 13 '13 at 1:15
2
2
You can add $q$ to $y$ as many times as needed to make it positive. Also, you can calculate the Legendre symbol value of $frac{-1}{q}$.
– Calvin Lin
May 13 '13 at 1:14
You can add $q$ to $y$ as many times as needed to make it positive. Also, you can calculate the Legendre symbol value of $frac{-1}{q}$.
– Calvin Lin
May 13 '13 at 1:14
$left( frac{-1}{q} right)=(-1)^frac{q-1}{2}$ for odd $q$... You would need to study separately what happens when $q$ is even and prime ;)
– N. S.
May 13 '13 at 1:15
$left( frac{-1}{q} right)=(-1)^frac{q-1}{2}$ for odd $q$... You would need to study separately what happens when $q$ is even and prime ;)
– N. S.
May 13 '13 at 1:15
add a comment |
3 Answers
3
active
oldest
votes
You can take out the $-1$ using the multiplicativity of the Legendre symbol, and use the first supplementary law:
$$left(frac{-a}{q}right)=left(frac{-1}{q}right)left(frac{a}{q}right)=(-1)^{(q-1)/2}left(frac{a}{q}right)$$
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
Gosh, I feel silly! Thank you!
– Damien
May 13 '13 at 1:33
add a comment |
There are $frac{p-1}{2}$ quadratic residues between $1$ and $p-1$. Let $x$ be any of these, and let $y=x-17p$. Since $x$ is a quadratic residue of $p$, so is $y$.
answered May 13 '13 at 4:16
André Nicolas
451k36421805
add a comment |
Yet another easy problem: For some integer $p$, pick any integer $x$ and compute $a=x^2pmod p$. $-p+a$ is a negative number that is a quadratic residue $pmod p$. In fact, $pk+a$ is a quadratic residue that is a negative number as long as $k$ is negative.
answered Nov 20 '18 at 1:57
J. Linne
846315
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can take out the $-1$ using the multiplicativity of the Legendre symbol, and use the first supplementary law:
$$left(frac{-a}{q}right)=left(frac{-1}{q}right)left(frac{a}{q}right)=(-1)^{(q-1)/2}left(frac{a}{q}right)$$
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
Gosh, I feel silly! Thank you!
– Damien
May 13 '13 at 1:33
add a comment |
You can take out the $-1$ using the multiplicativity of the Legendre symbol, and use the first supplementary law:
$$left(frac{-a}{q}right)=left(frac{-1}{q}right)left(frac{a}{q}right)=(-1)^{(q-1)/2}left(frac{a}{q}right)$$
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
Gosh, I feel silly! Thank you!
– Damien
May 13 '13 at 1:33
add a comment |
You can take out the $-1$ using the multiplicativity of the Legendre symbol, and use the first supplementary law:
$$left(frac{-a}{q}right)=left(frac{-1}{q}right)left(frac{a}{q}right)=(-1)^{(q-1)/2}left(frac{a}{q}right)$$
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
You can take out the $-1$ using the multiplicativity of the Legendre symbol, and use the first supplementary law:
$$left(frac{-a}{q}right)=left(frac{-1}{q}right)left(frac{a}{q}right)=(-1)^{(q-1)/2}left(frac{a}{q}right)$$
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
answered May 13 '13 at 1:14
Zev Chonoles
109k16225422
109k16225422
Gosh, I feel silly! Thank you!
– Damien
May 13 '13 at 1:33
add a comment |
Gosh, I feel silly! Thank you!
– Damien
May 13 '13 at 1:33
Gosh, I feel silly! Thank you!
– Damien
May 13 '13 at 1:33
Gosh, I feel silly! Thank you!
– Damien
May 13 '13 at 1:33
add a comment |
There are $frac{p-1}{2}$ quadratic residues between $1$ and $p-1$. Let $x$ be any of these, and let $y=x-17p$. Since $x$ is a quadratic residue of $p$, so is $y$.
answered May 13 '13 at 4:16
André Nicolas
451k36421805
add a comment |
There are $frac{p-1}{2}$ quadratic residues between $1$ and $p-1$. Let $x$ be any of these, and let $y=x-17p$. Since $x$ is a quadratic residue of $p$, so is $y$.
answered May 13 '13 at 4:16
André Nicolas
451k36421805
add a comment |
There are $frac{p-1}{2}$ quadratic residues between $1$ and $p-1$. Let $x$ be any of these, and let $y=x-17p$. Since $x$ is a quadratic residue of $p$, so is $y$.
answered May 13 '13 at 4:16
André Nicolas
451k36421805
There are $frac{p-1}{2}$ quadratic residues between $1$ and $p-1$. Let $x$ be any of these, and let $y=x-17p$. Since $x$ is a quadratic residue of $p$, so is $y$.
answered May 13 '13 at 4:16
André Nicolas
451k36421805
answered May 13 '13 at 4:16
André Nicolas
451k36421805
answered May 13 '13 at 4:16
André Nicolas
451k36421805
answered May 13 '13 at 4:16
André Nicolas
451k36421805
451k36421805
add a comment |
add a comment |
Yet another easy problem: For some integer $p$, pick any integer $x$ and compute $a=x^2pmod p$. $-p+a$ is a negative number that is a quadratic residue $pmod p$. In fact, $pk+a$ is a quadratic residue that is a negative number as long as $k$ is negative.
answered Nov 20 '18 at 1:57
J. Linne
846315
add a comment |
Yet another easy problem: For some integer $p$, pick any integer $x$ and compute $a=x^2pmod p$. $-p+a$ is a negative number that is a quadratic residue $pmod p$. In fact, $pk+a$ is a quadratic residue that is a negative number as long as $k$ is negative.
answered Nov 20 '18 at 1:57
J. Linne
846315
add a comment |
Yet another easy problem: For some integer $p$, pick any integer $x$ and compute $a=x^2pmod p$. $-p+a$ is a negative number that is a quadratic residue $pmod p$. In fact, $pk+a$ is a quadratic residue that is a negative number as long as $k$ is negative.
answered Nov 20 '18 at 1:57
J. Linne
846315
Yet another easy problem: For some integer $p$, pick any integer $x$ and compute $a=x^2pmod p$. $-p+a$ is a negative number that is a quadratic residue $pmod p$. In fact, $pk+a$ is a quadratic residue that is a negative number as long as $k$ is negative.
answered Nov 20 '18 at 1:57
J. Linne
846315
answered Nov 20 '18 at 1:57
J. Linne
846315
answered Nov 20 '18 at 1:57
J. Linne
846315
answered Nov 20 '18 at 1:57
J. Linne
846315
846315
add a comment |
add a comment |
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2
You can add $q$ to $y$ as many times as needed to make it positive. Also, you can calculate the Legendre symbol value of $frac{-1}{q}$.
– Calvin Lin
May 13 '13 at 1:14
$left( frac{-1}{q} right)=(-1)^frac{q-1}{2}$ for odd $q$... You would need to study separately what happens when $q$ is even and prime ;)
– N. S.
May 13 '13 at 1:15