If $p=a^2+b^2$ prove these consequences about $big(!frac{a}{p}!big)$
Suppose odd prime $p=a^2+b^2$, and $a$ is odd and $b$ is even. Prove that if $bequiv2pmod4$, then $left(dfrac bpright)=-1$ and if $bequiv0pmod4$, then $left(dfrac bpright)=1$.
What I have got is that $pequiv 1 pmod4$. I have tried factoring $b$ and splitting up the Legendre symbols. However, I am not sure this gets anywhere because you cannot say much about those. I have also observed that $b equiv 0 pmod 4$ when $pequiv 1 pmod 8$ and $b equiv 2 pmod 4$ when $pequiv 5 pmod 8$ if that helps.
number-theory prime-numbers legendre-symbol
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
add a comment |
Suppose odd prime $p=a^2+b^2$, and $a$ is odd and $b$ is even. Prove that if $bequiv2pmod4$, then $left(dfrac bpright)=-1$ and if $bequiv0pmod4$, then $left(dfrac bpright)=1$.
What I have got is that $pequiv 1 pmod4$. I have tried factoring $b$ and splitting up the Legendre symbols. However, I am not sure this gets anywhere because you cannot say much about those. I have also observed that $b equiv 0 pmod 4$ when $pequiv 1 pmod 8$ and $b equiv 2 pmod 4$ when $pequiv 5 pmod 8$ if that helps.
number-theory prime-numbers legendre-symbol
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
@twnly so "for each $q | b, p $ is a square $bmod q$, then use quadratic reciprocity on $prod_j (frac{q_j}{p})^{e_j}$
– reuns
Dec 1 '18 at 6:27
@reuns is right, my proposed solution has some errors. I deleted them.
– twnly
Dec 1 '18 at 6:27
is there some reason to expect there is no much simpler solution ?
– reuns
Dec 1 '18 at 6:31
add a comment |
Suppose odd prime $p=a^2+b^2$, and $a$ is odd and $b$ is even. Prove that if $bequiv2pmod4$, then $left(dfrac bpright)=-1$ and if $bequiv0pmod4$, then $left(dfrac bpright)=1$.
What I have got is that $pequiv 1 pmod4$. I have tried factoring $b$ and splitting up the Legendre symbols. However, I am not sure this gets anywhere because you cannot say much about those. I have also observed that $b equiv 0 pmod 4$ when $pequiv 1 pmod 8$ and $b equiv 2 pmod 4$ when $pequiv 5 pmod 8$ if that helps.
number-theory prime-numbers legendre-symbol
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
Suppose odd prime $p=a^2+b^2$, and $a$ is odd and $b$ is even. Prove that if $bequiv2pmod4$, then $left(dfrac bpright)=-1$ and if $bequiv0pmod4$, then $left(dfrac bpright)=1$.
What I have got is that $pequiv 1 pmod4$. I have tried factoring $b$ and splitting up the Legendre symbols. However, I am not sure this gets anywhere because you cannot say much about those. I have also observed that $b equiv 0 pmod 4$ when $pequiv 1 pmod 8$ and $b equiv 2 pmod 4$ when $pequiv 5 pmod 8$ if that helps.
number-theory prime-numbers legendre-symbol
number-theory prime-numbers legendre-symbol
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
edited Dec 1 '18 at 6:47
Henning Makholm
239k16303540
239k16303540
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
asked Dec 1 '18 at 5:07
Brian LiuBrian Liu
513
513
@twnly so "for each $q | b, p $ is a square $bmod q$, then use quadratic reciprocity on $prod_j (frac{q_j}{p})^{e_j}$
– reuns
Dec 1 '18 at 6:27
@reuns is right, my proposed solution has some errors. I deleted them.
– twnly
Dec 1 '18 at 6:27
is there some reason to expect there is no much simpler solution ?
– reuns
Dec 1 '18 at 6:31
add a comment |
@twnly so "for each $q | b, p $ is a square $bmod q$, then use quadratic reciprocity on $prod_j (frac{q_j}{p})^{e_j}$
– reuns
Dec 1 '18 at 6:27
@reuns is right, my proposed solution has some errors. I deleted them.
– twnly
Dec 1 '18 at 6:27
is there some reason to expect there is no much simpler solution ?
– reuns
Dec 1 '18 at 6:31
@twnly so "for each $q | b, p $ is a square $bmod q$, then use quadratic reciprocity on $prod_j (frac{q_j}{p})^{e_j}$
– reuns
Dec 1 '18 at 6:27
@twnly so "for each $q | b, p $ is a square $bmod q$, then use quadratic reciprocity on $prod_j (frac{q_j}{p})^{e_j}$
– reuns
Dec 1 '18 at 6:27
@reuns is right, my proposed solution has some errors. I deleted them.
– twnly
Dec 1 '18 at 6:27
@reuns is right, my proposed solution has some errors. I deleted them.
– twnly
Dec 1 '18 at 6:27
is there some reason to expect there is no much simpler solution ?
– reuns
Dec 1 '18 at 6:31
is there some reason to expect there is no much simpler solution ?
– reuns
Dec 1 '18 at 6:31
add a comment |
1 Answer
1
active
oldest
votes
Write $b=2^rc$ with $c$ odd. Then $p=a^2+b^2equiv a^2pmod c$ and so
$left(frac pcright)=1$ (this is a Jacobi symbol). As $pequiv1
pmod 4$ then $left(frac cpright)=1$ (quadratic reciprocity for
Jacobi symbols).
When $pequiv1pmod8$ then $left(frac 2pright)=1$ and so $left(frac bpright)=left(frac 2pright)^rleft(frac cpright)=1$. When $pequiv5pmod8$ then $left(frac 2pright)=-1$ and also
$r=1$. Therefore $left(frac bpright)=left(frac 2pright)left(frac cpright)=-1$.
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
add a comment |
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1 Answer
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1 Answer
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Write $b=2^rc$ with $c$ odd. Then $p=a^2+b^2equiv a^2pmod c$ and so
$left(frac pcright)=1$ (this is a Jacobi symbol). As $pequiv1
pmod 4$ then $left(frac cpright)=1$ (quadratic reciprocity for
Jacobi symbols).
When $pequiv1pmod8$ then $left(frac 2pright)=1$ and so $left(frac bpright)=left(frac 2pright)^rleft(frac cpright)=1$. When $pequiv5pmod8$ then $left(frac 2pright)=-1$ and also
$r=1$. Therefore $left(frac bpright)=left(frac 2pright)left(frac cpright)=-1$.
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
add a comment |
Write $b=2^rc$ with $c$ odd. Then $p=a^2+b^2equiv a^2pmod c$ and so
$left(frac pcright)=1$ (this is a Jacobi symbol). As $pequiv1
pmod 4$ then $left(frac cpright)=1$ (quadratic reciprocity for
Jacobi symbols).
When $pequiv1pmod8$ then $left(frac 2pright)=1$ and so $left(frac bpright)=left(frac 2pright)^rleft(frac cpright)=1$. When $pequiv5pmod8$ then $left(frac 2pright)=-1$ and also
$r=1$. Therefore $left(frac bpright)=left(frac 2pright)left(frac cpright)=-1$.
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
add a comment |
Write $b=2^rc$ with $c$ odd. Then $p=a^2+b^2equiv a^2pmod c$ and so
$left(frac pcright)=1$ (this is a Jacobi symbol). As $pequiv1
pmod 4$ then $left(frac cpright)=1$ (quadratic reciprocity for
Jacobi symbols).
When $pequiv1pmod8$ then $left(frac 2pright)=1$ and so $left(frac bpright)=left(frac 2pright)^rleft(frac cpright)=1$. When $pequiv5pmod8$ then $left(frac 2pright)=-1$ and also
$r=1$. Therefore $left(frac bpright)=left(frac 2pright)left(frac cpright)=-1$.
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
Write $b=2^rc$ with $c$ odd. Then $p=a^2+b^2equiv a^2pmod c$ and so
$left(frac pcright)=1$ (this is a Jacobi symbol). As $pequiv1
pmod 4$ then $left(frac cpright)=1$ (quadratic reciprocity for
Jacobi symbols).
When $pequiv1pmod8$ then $left(frac 2pright)=1$ and so $left(frac bpright)=left(frac 2pright)^rleft(frac cpright)=1$. When $pequiv5pmod8$ then $left(frac 2pright)=-1$ and also
$r=1$. Therefore $left(frac bpright)=left(frac 2pright)left(frac cpright)=-1$.
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Dec 1 '18 at 6:33
Lord Shark the UnknownLord Shark the Unknown
102k959132
102k959132
add a comment |
add a comment |
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@twnly so "for each $q | b, p $ is a square $bmod q$, then use quadratic reciprocity on $prod_j (frac{q_j}{p})^{e_j}$
– reuns
Dec 1 '18 at 6:27
@reuns is right, my proposed solution has some errors. I deleted them.
– twnly
Dec 1 '18 at 6:27
is there some reason to expect there is no much simpler solution ?
– reuns
Dec 1 '18 at 6:31