Determine the Legendre symbol of $left(frac{14}{p}right)$
$begingroup$
I have been asked to determine the Legendre symbol $left(frac{14}{p}right)$ for $p geq 11$ and have made good progress, however, I am stuck at the very last hurdle. So far, I have found that
begin{equation}
left(frac{14}{p}right) = left(frac{2}{p}right)left(frac{7}{p}right)
end{equation}
and we know that
begin{equation}
left(frac{2}{p}right) = (-1)^{frac{p^2 - 1}{8}} =begin{cases}
1, & p equiv pm 1 ;bmod; 8\
-1, & p equiv pm 3 ;bmod; 8
end{cases}
end{equation}
and I have calculated that
begin{equation}
left(frac{7}{p}right) =begin{cases}
1, & p equiv pm 1, pm 3, pm 9 ;bmod; 28\
-1, & p equiv pm 5, pm 11, pm 13 ;bmod; 28.
end{cases}
end{equation}
I am attempting to summarise these using the Chinese Remainder Theorem, however, I am having no luck. Using the Chinese Remainder Theorem similar to the way I used it when calculating $left(frac{7}{p}right)$ would just be too many cases to consider. Am I not spotting an easy method to combine these Legendre Symbols?
This question is similar to that which was posted here, but there are no relevant answers to that post.
number-theory elementary-number-theory modular-arithmetic legendre-symbol
asked Dec 5 '18 at 10:19
YGradeYGrade
256
$endgroup$
add a comment |
$begingroup$
I have been asked to determine the Legendre symbol $left(frac{14}{p}right)$ for $p geq 11$ and have made good progress, however, I am stuck at the very last hurdle. So far, I have found that
begin{equation}
left(frac{14}{p}right) = left(frac{2}{p}right)left(frac{7}{p}right)
end{equation}
and we know that
begin{equation}
left(frac{2}{p}right) = (-1)^{frac{p^2 - 1}{8}} =begin{cases}
1, & p equiv pm 1 ;bmod; 8\
-1, & p equiv pm 3 ;bmod; 8
end{cases}
end{equation}
and I have calculated that
begin{equation}
left(frac{7}{p}right) =begin{cases}
1, & p equiv pm 1, pm 3, pm 9 ;bmod; 28\
-1, & p equiv pm 5, pm 11, pm 13 ;bmod; 28.
end{cases}
end{equation}
I am attempting to summarise these using the Chinese Remainder Theorem, however, I am having no luck. Using the Chinese Remainder Theorem similar to the way I used it when calculating $left(frac{7}{p}right)$ would just be too many cases to consider. Am I not spotting an easy method to combine these Legendre Symbols?
This question is similar to that which was posted here, but there are no relevant answers to that post.
number-theory elementary-number-theory modular-arithmetic legendre-symbol
asked Dec 5 '18 at 10:19
YGradeYGrade
256
$endgroup$
$begingroup$
You'll need to work modulo $56$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 10:30
$begingroup$
@LordSharktheUnknown any chance you could expand on the method of converting my working to modulo 56?
$endgroup$
– YGrade
Dec 5 '18 at 10:50
2
$begingroup$
$x equiv 1 pmod {28}$ means $x equiv 1 pmod {56}$ or $x equiv 29 pmod {56}$. Similiarly $x equiv -3 pmod 8$ means $x equiv 5 pmod {56}$ or $x equiv 13 pmod {56}$, or $ldots$, or $x equiv 29 pmod {56}$, or $ldots$, or $x equiv 53 pmod {56}$. You now know the value of both partial Legendre symbols for $p equiv 29 pmod {56}$, and can finally conculde $left(frac{14}pright)$ for $p equiv 29 pmod {56}$. Continue this for all the other residue classes. A simple (though slightly long) table will help, I don't think using CRT will be less work in this case.
$endgroup$
– Ingix
Dec 5 '18 at 11:24
add a comment |
$begingroup$
I have been asked to determine the Legendre symbol $left(frac{14}{p}right)$ for $p geq 11$ and have made good progress, however, I am stuck at the very last hurdle. So far, I have found that
begin{equation}
left(frac{14}{p}right) = left(frac{2}{p}right)left(frac{7}{p}right)
end{equation}
and we know that
begin{equation}
left(frac{2}{p}right) = (-1)^{frac{p^2 - 1}{8}} =begin{cases}
1, & p equiv pm 1 ;bmod; 8\
-1, & p equiv pm 3 ;bmod; 8
end{cases}
end{equation}
and I have calculated that
begin{equation}
left(frac{7}{p}right) =begin{cases}
1, & p equiv pm 1, pm 3, pm 9 ;bmod; 28\
-1, & p equiv pm 5, pm 11, pm 13 ;bmod; 28.
end{cases}
end{equation}
I am attempting to summarise these using the Chinese Remainder Theorem, however, I am having no luck. Using the Chinese Remainder Theorem similar to the way I used it when calculating $left(frac{7}{p}right)$ would just be too many cases to consider. Am I not spotting an easy method to combine these Legendre Symbols?
This question is similar to that which was posted here, but there are no relevant answers to that post.
number-theory elementary-number-theory modular-arithmetic legendre-symbol
asked Dec 5 '18 at 10:19
YGradeYGrade
256
$endgroup$
I have been asked to determine the Legendre symbol $left(frac{14}{p}right)$ for $p geq 11$ and have made good progress, however, I am stuck at the very last hurdle. So far, I have found that
begin{equation}
left(frac{14}{p}right) = left(frac{2}{p}right)left(frac{7}{p}right)
end{equation}
and we know that
begin{equation}
left(frac{2}{p}right) = (-1)^{frac{p^2 - 1}{8}} =begin{cases}
1, & p equiv pm 1 ;bmod; 8\
-1, & p equiv pm 3 ;bmod; 8
end{cases}
end{equation}
and I have calculated that
begin{equation}
left(frac{7}{p}right) =begin{cases}
1, & p equiv pm 1, pm 3, pm 9 ;bmod; 28\
-1, & p equiv pm 5, pm 11, pm 13 ;bmod; 28.
end{cases}
end{equation}
I am attempting to summarise these using the Chinese Remainder Theorem, however, I am having no luck. Using the Chinese Remainder Theorem similar to the way I used it when calculating $left(frac{7}{p}right)$ would just be too many cases to consider. Am I not spotting an easy method to combine these Legendre Symbols?
This question is similar to that which was posted here, but there are no relevant answers to that post.
number-theory elementary-number-theory modular-arithmetic legendre-symbol
number-theory elementary-number-theory modular-arithmetic legendre-symbol
asked Dec 5 '18 at 10:19
YGradeYGrade
256
asked Dec 5 '18 at 10:19
YGradeYGrade
256
asked Dec 5 '18 at 10:19
YGradeYGrade
256
asked Dec 5 '18 at 10:19
YGradeYGrade
256
asked Dec 5 '18 at 10:19
YGradeYGrade
256
256
$begingroup$
You'll need to work modulo $56$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 10:30
$begingroup$
@LordSharktheUnknown any chance you could expand on the method of converting my working to modulo 56?
$endgroup$
– YGrade
Dec 5 '18 at 10:50
2
$begingroup$
$x equiv 1 pmod {28}$ means $x equiv 1 pmod {56}$ or $x equiv 29 pmod {56}$. Similiarly $x equiv -3 pmod 8$ means $x equiv 5 pmod {56}$ or $x equiv 13 pmod {56}$, or $ldots$, or $x equiv 29 pmod {56}$, or $ldots$, or $x equiv 53 pmod {56}$. You now know the value of both partial Legendre symbols for $p equiv 29 pmod {56}$, and can finally conculde $left(frac{14}pright)$ for $p equiv 29 pmod {56}$. Continue this for all the other residue classes. A simple (though slightly long) table will help, I don't think using CRT will be less work in this case.
$endgroup$
– Ingix
Dec 5 '18 at 11:24
add a comment |
$begingroup$
You'll need to work modulo $56$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 10:30
$begingroup$
@LordSharktheUnknown any chance you could expand on the method of converting my working to modulo 56?
$endgroup$
– YGrade
Dec 5 '18 at 10:50
2
$begingroup$
$x equiv 1 pmod {28}$ means $x equiv 1 pmod {56}$ or $x equiv 29 pmod {56}$. Similiarly $x equiv -3 pmod 8$ means $x equiv 5 pmod {56}$ or $x equiv 13 pmod {56}$, or $ldots$, or $x equiv 29 pmod {56}$, or $ldots$, or $x equiv 53 pmod {56}$. You now know the value of both partial Legendre symbols for $p equiv 29 pmod {56}$, and can finally conculde $left(frac{14}pright)$ for $p equiv 29 pmod {56}$. Continue this for all the other residue classes. A simple (though slightly long) table will help, I don't think using CRT will be less work in this case.
$endgroup$
– Ingix
Dec 5 '18 at 11:24
$begingroup$
You'll need to work modulo $56$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 10:30
$begingroup$
You'll need to work modulo $56$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 10:30
$begingroup$
@LordSharktheUnknown any chance you could expand on the method of converting my working to modulo 56?
$endgroup$
– YGrade
Dec 5 '18 at 10:50
$begingroup$
@LordSharktheUnknown any chance you could expand on the method of converting my working to modulo 56?
$endgroup$
– YGrade
Dec 5 '18 at 10:50
2
2
$begingroup$
$x equiv 1 pmod {28}$ means $x equiv 1 pmod {56}$ or $x equiv 29 pmod {56}$. Similiarly $x equiv -3 pmod 8$ means $x equiv 5 pmod {56}$ or $x equiv 13 pmod {56}$, or $ldots$, or $x equiv 29 pmod {56}$, or $ldots$, or $x equiv 53 pmod {56}$. You now know the value of both partial Legendre symbols for $p equiv 29 pmod {56}$, and can finally conculde $left(frac{14}pright)$ for $p equiv 29 pmod {56}$. Continue this for all the other residue classes. A simple (though slightly long) table will help, I don't think using CRT will be less work in this case.
$endgroup$
– Ingix
Dec 5 '18 at 11:24
$begingroup$
$x equiv 1 pmod {28}$ means $x equiv 1 pmod {56}$ or $x equiv 29 pmod {56}$. Similiarly $x equiv -3 pmod 8$ means $x equiv 5 pmod {56}$ or $x equiv 13 pmod {56}$, or $ldots$, or $x equiv 29 pmod {56}$, or $ldots$, or $x equiv 53 pmod {56}$. You now know the value of both partial Legendre symbols for $p equiv 29 pmod {56}$, and can finally conculde $left(frac{14}pright)$ for $p equiv 29 pmod {56}$. Continue this for all the other residue classes. A simple (though slightly long) table will help, I don't think using CRT will be less work in this case.
$endgroup$
– Ingix
Dec 5 '18 at 11:24
add a comment |
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$begingroup$
You'll need to work modulo $56$.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 10:30
$begingroup$
@LordSharktheUnknown any chance you could expand on the method of converting my working to modulo 56?
$endgroup$
– YGrade
Dec 5 '18 at 10:50
2
$begingroup$
$x equiv 1 pmod {28}$ means $x equiv 1 pmod {56}$ or $x equiv 29 pmod {56}$. Similiarly $x equiv -3 pmod 8$ means $x equiv 5 pmod {56}$ or $x equiv 13 pmod {56}$, or $ldots$, or $x equiv 29 pmod {56}$, or $ldots$, or $x equiv 53 pmod {56}$. You now know the value of both partial Legendre symbols for $p equiv 29 pmod {56}$, and can finally conculde $left(frac{14}pright)$ for $p equiv 29 pmod {56}$. Continue this for all the other residue classes. A simple (though slightly long) table will help, I don't think using CRT will be less work in this case.
$endgroup$
– Ingix
Dec 5 '18 at 11:24