the solutions are quadratic nonresidues modulo $p$

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Let $q$ be an odd prime number, such that $p=4q+1$ is also an odd prime. I want to show that the congruence $x^2 equiv -1 pmodp$ has two (incongruenct modulo $p$) solutions, which are quadratic nonresidues modulo $p$.



I have thought to consider the Jacobi symbol.



We have that $left( frac-1pright)=left( frac-14q+1right)=(-1)^frac4q+1-12=(-1)^2q=1$.



From this we deduce that $x^2 equiv -1 pmodp$ has two incongruent modulo $p$ solutions, right?



How do we deduce that the solutions are quadratic nonresidues modulo $p$ ?




number-theory quadratic-residues




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    Hint: suppose you got quadratic residues. Deduce the existence of an element of order $8$, derive contradiction.
    – lulu
    Aug 6 at 20:14















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Let $q$ be an odd prime number, such that $p=4q+1$ is also an odd prime. I want to show that the congruence $x^2 equiv -1 pmodp$ has two (incongruenct modulo $p$) solutions, which are quadratic nonresidues modulo $p$.



I have thought to consider the Jacobi symbol.



We have that $left( frac-1pright)=left( frac-14q+1right)=(-1)^frac4q+1-12=(-1)^2q=1$.



From this we deduce that $x^2 equiv -1 pmodp$ has two incongruent modulo $p$ solutions, right?



How do we deduce that the solutions are quadratic nonresidues modulo $p$ ?




number-theory quadratic-residues




share|cite|improve this question















  • 1




    Hint: suppose you got quadratic residues. Deduce the existence of an element of order $8$, derive contradiction.
    – lulu
    Aug 6 at 20:14













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $q$ be an odd prime number, such that $p=4q+1$ is also an odd prime. I want to show that the congruence $x^2 equiv -1 pmodp$ has two (incongruenct modulo $p$) solutions, which are quadratic nonresidues modulo $p$.



I have thought to consider the Jacobi symbol.



We have that $left( frac-1pright)=left( frac-14q+1right)=(-1)^frac4q+1-12=(-1)^2q=1$.



From this we deduce that $x^2 equiv -1 pmodp$ has two incongruent modulo $p$ solutions, right?



How do we deduce that the solutions are quadratic nonresidues modulo $p$ ?




number-theory quadratic-residues




share|cite|improve this question











Let $q$ be an odd prime number, such that $p=4q+1$ is also an odd prime. I want to show that the congruence $x^2 equiv -1 pmodp$ has two (incongruenct modulo $p$) solutions, which are quadratic nonresidues modulo $p$.



I have thought to consider the Jacobi symbol.



We have that $left( frac-1pright)=left( frac-14q+1right)=(-1)^frac4q+1-12=(-1)^2q=1$.



From this we deduce that $x^2 equiv -1 pmodp$ has two incongruent modulo $p$ solutions, right?



How do we deduce that the solutions are quadratic nonresidues modulo $p$ ?





number-theory quadratic-residues





share|cite|improve this question










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asked Aug 6 at 20:09









Evinda

451412




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  • 1




    Hint: suppose you got quadratic residues. Deduce the existence of an element of order $8$, derive contradiction.
    – lulu
    Aug 6 at 20:14













  • 1




    Hint: suppose you got quadratic residues. Deduce the existence of an element of order $8$, derive contradiction.
    – lulu
    Aug 6 at 20:14








1




1




Hint: suppose you got quadratic residues. Deduce the existence of an element of order $8$, derive contradiction.
– lulu
Aug 6 at 20:14





Hint: suppose you got quadratic residues. Deduce the existence of an element of order $8$, derive contradiction.
– lulu
Aug 6 at 20:14











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Suppose $x$ could be a quadratic residue modulo $p$ then we could write $xequiv a^2bmod p$ for some integer $a$ therefore $a^4equiv -1bmod pimplies a^p-1equiv (-1)^fracp-14bmod pimplies 1equiv (-1)^qbmod p$ but since $q$ is an odd prime this means $1equiv -1bmod pimplies 2equiv 0bmod pimplies pmid 2implies p=2$ which is a contradiction. Therefore $x$ must be quadratic non-residue or equivilently $left( fracxpright)=-1$, and since $left( frac-xpright)=left( frac-1pright)left( fracxpright)=left( fracxpright)=-1$ we see $-x$ is also a quadratic non-residue, thus all solutions $y$ to the equation $y^2equiv -1bmod p$ are quadratic non-residues modulo $p$.






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    Suppose $x$ could be a quadratic residue modulo $p$ then we could write $xequiv a^2bmod p$ for some integer $a$ therefore $a^4equiv -1bmod pimplies a^p-1equiv (-1)^fracp-14bmod pimplies 1equiv (-1)^qbmod p$ but since $q$ is an odd prime this means $1equiv -1bmod pimplies 2equiv 0bmod pimplies pmid 2implies p=2$ which is a contradiction. Therefore $x$ must be quadratic non-residue or equivilently $left( fracxpright)=-1$, and since $left( frac-xpright)=left( frac-1pright)left( fracxpright)=left( fracxpright)=-1$ we see $-x$ is also a quadratic non-residue, thus all solutions $y$ to the equation $y^2equiv -1bmod p$ are quadratic non-residues modulo $p$.






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      Suppose $x$ could be a quadratic residue modulo $p$ then we could write $xequiv a^2bmod p$ for some integer $a$ therefore $a^4equiv -1bmod pimplies a^p-1equiv (-1)^fracp-14bmod pimplies 1equiv (-1)^qbmod p$ but since $q$ is an odd prime this means $1equiv -1bmod pimplies 2equiv 0bmod pimplies pmid 2implies p=2$ which is a contradiction. Therefore $x$ must be quadratic non-residue or equivilently $left( fracxpright)=-1$, and since $left( frac-xpright)=left( frac-1pright)left( fracxpright)=left( fracxpright)=-1$ we see $-x$ is also a quadratic non-residue, thus all solutions $y$ to the equation $y^2equiv -1bmod p$ are quadratic non-residues modulo $p$.






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        up vote
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        up vote
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        Suppose $x$ could be a quadratic residue modulo $p$ then we could write $xequiv a^2bmod p$ for some integer $a$ therefore $a^4equiv -1bmod pimplies a^p-1equiv (-1)^fracp-14bmod pimplies 1equiv (-1)^qbmod p$ but since $q$ is an odd prime this means $1equiv -1bmod pimplies 2equiv 0bmod pimplies pmid 2implies p=2$ which is a contradiction. Therefore $x$ must be quadratic non-residue or equivilently $left( fracxpright)=-1$, and since $left( frac-xpright)=left( frac-1pright)left( fracxpright)=left( fracxpright)=-1$ we see $-x$ is also a quadratic non-residue, thus all solutions $y$ to the equation $y^2equiv -1bmod p$ are quadratic non-residues modulo $p$.






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        Suppose $x$ could be a quadratic residue modulo $p$ then we could write $xequiv a^2bmod p$ for some integer $a$ therefore $a^4equiv -1bmod pimplies a^p-1equiv (-1)^fracp-14bmod pimplies 1equiv (-1)^qbmod p$ but since $q$ is an odd prime this means $1equiv -1bmod pimplies 2equiv 0bmod pimplies pmid 2implies p=2$ which is a contradiction. Therefore $x$ must be quadratic non-residue or equivilently $left( fracxpright)=-1$, and since $left( frac-xpright)=left( frac-1pright)left( fracxpright)=left( fracxpright)=-1$ we see $-x$ is also a quadratic non-residue, thus all solutions $y$ to the equation $y^2equiv -1bmod p$ are quadratic non-residues modulo $p$.







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        answered Aug 6 at 20:20









        Ethan

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