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Legendre Symbol: show that $left(frac{-3}pright)=left(frac p3right)$

Legendre symbol: Showing that $sum_{m=0}^{p-1} left(frac{am+b}{p}right)=0$Legendre symbol $left(frac 3pright)$Prove that Legendre Symbol $left(frac{(p-1)!}pright) equiv ppmod4$Legendre Symbol $left(frac4pright)$ is always congruent to $1$?Proving summations involving the Legendre symbolLegendre symbol $(-21/p)$What is the solution to $Fleft(nright)=Fleft(n-1right)+left(frac{2}{n}right)Fleft(n-2right)$ with Legendre symbol?Sum of Legendre symbol $left(frac{n^2-a}{p}right)$ (More explanation)Determine the Legendre symbol of $left(frac{14}{p}right)$Characterization for the Legendre Symbol

1

$begingroup$

I am asked to solve the following question

Let $p$ be a prime, $p>3$. By considering the cases $p ≡1bmod4$ and $p≡-1bmod4$ separately, show that $left(frac{-3}pright)=left(frac p3right)$.

So if $p ≡ 1bmod4$, $left(frac{-3}pright)=left(frac p{-3}right)$ and if $p ≡ -1bmod4$, $left(frac{-3}pright)=-left(frac p{-3}right)$.

And then I get stuck on how to process on; is there anything I can do to deal with the minus sign of $3$?

Thanks

elementary-number-theory legendre-symbol

share|cite|improve this question

edited Mar 10 at 3:46

Parcly Taxel

44.3k1375107

asked Mar 10 at 3:07

ThomasThomas

636

$endgroup$

add a comment | 

1

$begingroup$

I am asked to solve the following question

Let $p$ be a prime, $p>3$. By considering the cases $p ≡1bmod4$ and $p≡-1bmod4$ separately, show that $left(frac{-3}pright)=left(frac p3right)$.

So if $p ≡ 1bmod4$, $left(frac{-3}pright)=left(frac p{-3}right)$ and if $p ≡ -1bmod4$, $left(frac{-3}pright)=-left(frac p{-3}right)$.

And then I get stuck on how to process on; is there anything I can do to deal with the minus sign of $3$?

Thanks

elementary-number-theory legendre-symbol

share|cite|improve this question

edited Mar 10 at 3:46

Parcly Taxel

44.3k1375107

asked Mar 10 at 3:07

ThomasThomas

636

$endgroup$

add a comment | 

$begingroup$

I am asked to solve the following question

Let $p$ be a prime, $p>3$. By considering the cases $p ≡1bmod4$ and $p≡-1bmod4$ separately, show that $left(frac{-3}pright)=left(frac p3right)$.

So if $p ≡ 1bmod4$, $left(frac{-3}pright)=left(frac p{-3}right)$ and if $p ≡ -1bmod4$, $left(frac{-3}pright)=-left(frac p{-3}right)$.

And then I get stuck on how to process on; is there anything I can do to deal with the minus sign of $3$?

Thanks

elementary-number-theory legendre-symbol

share|cite|improve this question

edited Mar 10 at 3:46

Parcly Taxel

44.3k1375107

asked Mar 10 at 3:07

ThomasThomas

636

$endgroup$

share|cite|improve this question

edited Mar 10 at 3:46

Parcly Taxel

44.3k1375107

asked Mar 10 at 3:07

ThomasThomas

636

share|cite|improve this question

edited Mar 10 at 3:46

Parcly Taxel

44.3k1375107

asked Mar 10 at 3:07

ThomasThomas

636

edited Mar 10 at 3:46

Parcly Taxel

44.3k1375107

edited Mar 10 at 3:46

Parcly Taxel

44.3k1375107

asked Mar 10 at 3:07

ThomasThomas

636

asked Mar 10 at 3:07

ThomasThomas

636

1 Answer
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1

$begingroup$

Hint: Use the fact that $-1$ is a quadratic residue mod $p$ if and only if $pnotequiv3pmod{4}$.

share|cite|improve this answer

answered Mar 10 at 3:12

ServaesServaes

28.3k34099

$endgroup$

add a comment | 

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1

$begingroup$

Hint: Use the fact that $-1$ is a quadratic residue mod $p$ if and only if $pnotequiv3pmod{4}$.

share|cite|improve this answer

answered Mar 10 at 3:12

ServaesServaes

28.3k34099

$endgroup$

add a comment | 

1

$begingroup$

Hint: Use the fact that $-1$ is a quadratic residue mod $p$ if and only if $pnotequiv3pmod{4}$.

share|cite|improve this answer

answered Mar 10 at 3:12

ServaesServaes

28.3k34099

$endgroup$

add a comment | 

$begingroup$

Hint: Use the fact that $-1$ is a quadratic residue mod $p$ if and only if $pnotequiv3pmod{4}$.

share|cite|improve this answer

answered Mar 10 at 3:12

ServaesServaes

28.3k34099

$endgroup$

share|cite|improve this answer

answered Mar 10 at 3:12

ServaesServaes

28.3k34099

answered Mar 10 at 3:12

ServaesServaes

28.3k34099

answered Mar 10 at 3:12

ServaesServaes

28.3k34099