$n$th power residue conjecture











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Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.



Here is an example of the conjecture for $n=5$:



$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$



$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus



$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.



Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.










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  • 2




    Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
    – KCd
    yesterday












  • @KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
    – J. Linne
    22 hours ago










  • For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
    – KCd
    18 hours ago

















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Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.



Here is an example of the conjecture for $n=5$:



$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$



$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus



$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.



Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.










share|cite|improve this question




















  • 2




    Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
    – KCd
    yesterday












  • @KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
    – J. Linne
    22 hours ago










  • For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
    – KCd
    18 hours ago















up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.



Here is an example of the conjecture for $n=5$:



$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$



$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus



$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.



Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.










share|cite|improve this question















Let $left(dfrac{a}{p}right)_n$ be the $n$th power residue symbol such that when $a$ is an $n$th power residue $pmod p$, $left(dfrac{a}{p}right)_n=1$ otherwise $left(dfrac{a}{p}right)_n=zeta$. Let $f_n(x)$ be a defining polynomial for the $n$th cyclotomic field $K=mathbb{Q}(e^{2pi i/n})$. Suppose that there exists some integer $m$ such that $f_n(m)=0 pmod {n^2}$, and there are primes $p$ and $q$ such that $p=f_n(z)$ and $q=f_n(z+a)$. Then $left(dfrac{a}{p}right)_n = left(dfrac{a}{q}right)_n$. When $n=3$, this conjecture agrees with cubic reciprocity theorems. The same is true for $n=4$.



Here is an example of the conjecture for $n=5$:



$f_5(x)=x^4+5x^3+10x^2+25$ and $f_5(0)=0pmod {5^2}$



$a=3$, $p=f_5(1)=41$ and $q=f_5(1+a)=f_5(4)=761$ are both prime, thus



$left(dfrac{3}{41}right)_5 = left(dfrac{3}{761}right)_5$.



Since $3$ is a quintic residue $pmod {41}$, then it is also a quintic residue of $761$. In general, if $p=f_5(3a+1)$ is prime, then $3$ is a quintic residue $pmod p$.







elementary-number-theory cubic-reciprocity






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edited 22 hours ago

























asked 2 days ago









J. Linne

829315




829315








  • 2




    Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
    – KCd
    yesterday












  • @KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
    – J. Linne
    22 hours ago










  • For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
    – KCd
    18 hours ago
















  • 2




    Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
    – KCd
    yesterday












  • @KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
    – J. Linne
    22 hours ago










  • For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
    – KCd
    18 hours ago










2




2




Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
yesterday






Your $n$-th power residue symbol is not standard when $n > 2$: its value at numbers that are not $n$th powers mod $p$ should not be $-1$, but some $n$th root of unity if you want the symbol to be multiplicative in any reasonable way. Also, using ordinary prime numbers rather than prime ideals in $mathbf Z[e^{2pi i/n}]$ seems unusual, e.g., cubic reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/3}]$ and quartic (biquadratic) reciprocity is usually expressed in terms of $mathbf Z[e^{2pi i/4}] = mathbf Z[i]$. Your conjecture has a gap: what type of number is $z$?
– KCd
yesterday














@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
22 hours ago




@KCd changing -1 to a root of unity is only nessesary when dealing with cosets of the groups of $n$th power residues but I changed it anyways. As for $z$, like $x, a,$ and $n$ is an integer.
– J. Linne
22 hours ago












For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
18 hours ago






For $n > 2$, an $n$th power residue symbol does not just have two nonzero values, "1 and $zeta$". Its nonzero values run over all $n$th roots of unity, in order to be multiplicative in $a$, so your description of $(frac{a}{p})_n$ is unusual (esp. since $p$ need not be prime in $mathbf Z[e^{2pi i/n}]$, e.g., $p = 41$ when $n = 5$. Where are you getting your description of power residue symbols from?
– KCd
18 hours ago

















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