Proof that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime iif $p equiv pm 1 mod 10$












3












$begingroup$


Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28
















3












$begingroup$


Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28














3












3








3





$begingroup$


Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol







share|cite|improve this question











$endgroup$




Here I present the following proof in order to receive corrections or any kind of suggestion to improve my handling/knowledge of modular arithmetic:




Prove that $5$ is a quadratic residue $(mod p)$ with $p$ odd prime
iif $p equiv pm 1 mod 10$ ; prove also that $5$ is NOT a quadratic residue $(mod p)$
iif $p equiv pm 3 mod 10$.




Dim:



To check if $5$ is a quadratic residue $(mod p)$ I write the equivalent Legendre symbol with the condition:



$(5/p) = 1$



So I have for quadratic reciprocity $(5/p) = (p/5)(-1)^{{(p-1)(5-1)}over 4}=(p/5)(-1)^{(p-1)}$



$bullet$ The exponent $(p-1)$ must be $(mod2)$



$bullet$ $(p/5)$ means to find $p$ : $p(mod5)$ $rightarrow$ the choices are $1,3(mod5)$ because $p$ is prime



The moduli are coprime $(2,5)=1$ so I can study for the two final cases $(mod5times 2)=(mod10)$



Case $1(mod10)$:



Here $(1/5)=1$ and for the exponent $p=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=1$ for $p=1(mod10)$ but also for $p=-1(mod10)$



Case $3(mod10)$:



Here $(3/5)=-1$ because it's not a quadratic residue, and for the exponent $p=3(mod2)=1(mod2)$ so the exponent $(p-1)$ must be even. So $(5/p)=(-1)(1)=-1$ for $p=3(mod10)$ but also for $p=-3(mod10)$



$Box$



I appreciate any kind of critics and corrections.



Thank you






elementary-number-theory modular-arithmetic legendre-symbol




elementary-number-theory modular-arithmetic legendre-symbol






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 15:17







Alessar

















asked Nov 29 '18 at 14:41









AlessarAlessar

298115




298115








  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28














  • 1




    $begingroup$
    There are more cases. A prime can be $2$ or $4 pmod{5}.$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 12:42










  • $begingroup$
    I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
    $endgroup$
    – Alessar
    Nov 30 '18 at 13:08






  • 1




    $begingroup$
    Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
    $endgroup$
    – B. Goddard
    Nov 30 '18 at 13:31










  • $begingroup$
    Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
    $endgroup$
    – Alessar
    Dec 1 '18 at 11:28








1




1




$begingroup$
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
$endgroup$
– B. Goddard
Nov 30 '18 at 12:42




$begingroup$
There are more cases. A prime can be $2$ or $4 pmod{5}.$.
$endgroup$
– B. Goddard
Nov 30 '18 at 12:42












$begingroup$
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
$endgroup$
– Alessar
Nov 30 '18 at 13:08




$begingroup$
I've excluded 2 and 4 because p must be prime and odd; is it formally wrong to exclude them in this way a-priori?
$endgroup$
– Alessar
Nov 30 '18 at 13:08




1




1




$begingroup$
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
$endgroup$
– B. Goddard
Nov 30 '18 at 13:31




$begingroup$
Notice that $7 equiv 2 pmod{5}$ and $19equiv 4 pmod{5}$.
$endgroup$
– B. Goddard
Nov 30 '18 at 13:31












$begingroup$
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
$endgroup$
– Alessar
Dec 1 '18 at 11:28




$begingroup$
Thanks for the tip, I'll review the proof with your suggestions, I hope to edit it for the final version
$endgroup$
– Alessar
Dec 1 '18 at 11:28










1 Answer
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$begingroup$

See my other post use Gauss lemma to find $(frac{n}{p})$:



$(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






share|cite|improve this answer









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    active

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    1












    $begingroup$

    See my other post use Gauss lemma to find $(frac{n}{p})$:



    $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      See my other post use Gauss lemma to find $(frac{n}{p})$:



      $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        See my other post use Gauss lemma to find $(frac{n}{p})$:



        $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$






        share|cite|improve this answer









        $endgroup$



        See my other post use Gauss lemma to find $(frac{n}{p})$:



        $(frac{5}{p}) = 1 iff p equiv pm 1 mod 5$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 11:18









        Maestro13Maestro13

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