Legendre symbol $(-3/p)$ where $p = 1 mod 3$












1












$begingroup$


Suppose $p = 1 bmod 3$, prove the following statements:




  1. prove that $x^2 + x + 1 = 0 mod p$ has a solution

  2. Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$

  3. Determine the discriminant of $x^2 + x + 1$

  4. Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$


This is what I've tried by each question:




  1. prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.

  2. Note the following:
    $(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p.

  3. -3

  4. ?






elementary-number-theory quadratic-residues legendre-symbol







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
    $endgroup$
    – Jack D'Aurizio
    Oct 28 '15 at 14:25






  • 1




    $begingroup$
    1) makes no sense as written. Also such "question dumps" are not really welcome.
    $endgroup$
    – quid♦
    Oct 28 '15 at 14:30






  • 1




    $begingroup$
    What is $x$? Is it solve for $x$?
    $endgroup$
    – SchrodingersCat
    Oct 28 '15 at 14:33












  • $begingroup$
    sorry forgot the sentence has a solution i will edit it right away
    $endgroup$
    – Kees Til
    Oct 28 '15 at 14:36






  • 2




    $begingroup$
    The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
    $endgroup$
    – André Nicolas
    Oct 28 '15 at 16:38
















1












$begingroup$


Suppose $p = 1 bmod 3$, prove the following statements:




  1. prove that $x^2 + x + 1 = 0 mod p$ has a solution

  2. Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$

  3. Determine the discriminant of $x^2 + x + 1$

  4. Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$


This is what I've tried by each question:




  1. prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.

  2. Note the following:
    $(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p.

  3. -3

  4. ?






elementary-number-theory quadratic-residues legendre-symbol







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
    $endgroup$
    – Jack D'Aurizio
    Oct 28 '15 at 14:25






  • 1




    $begingroup$
    1) makes no sense as written. Also such "question dumps" are not really welcome.
    $endgroup$
    – quid♦
    Oct 28 '15 at 14:30






  • 1




    $begingroup$
    What is $x$? Is it solve for $x$?
    $endgroup$
    – SchrodingersCat
    Oct 28 '15 at 14:33












  • $begingroup$
    sorry forgot the sentence has a solution i will edit it right away
    $endgroup$
    – Kees Til
    Oct 28 '15 at 14:36






  • 2




    $begingroup$
    The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
    $endgroup$
    – André Nicolas
    Oct 28 '15 at 16:38














1












1








1


2



$begingroup$


Suppose $p = 1 bmod 3$, prove the following statements:




  1. prove that $x^2 + x + 1 = 0 mod p$ has a solution

  2. Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$

  3. Determine the discriminant of $x^2 + x + 1$

  4. Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$


This is what I've tried by each question:




  1. prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.

  2. Note the following:
    $(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p.

  3. -3

  4. ?






elementary-number-theory quadratic-residues legendre-symbol







share|cite|improve this question











$endgroup$




Suppose $p = 1 bmod 3$, prove the following statements:




  1. prove that $x^2 + x + 1 = 0 mod p$ has a solution

  2. Prove that $left(frac{-3}{p}right) = 1$ if $p = 1mod 3$

  3. Determine the discriminant of $x^2 + x + 1$

  4. Prove using 2,3 that $left(frac{-3}{p}right) = -1$ if $p = -1mod 3$


This is what I've tried by each question:




  1. prove $x^2 + x = -1 mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{frac{p-1}{2}} = -1 mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.

  2. Note the following:
    $(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p.

  3. -3

  4. ?






elementary-number-theory quadratic-residues legendre-symbol




elementary-number-theory quadratic-residues legendre-symbol






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 28 '15 at 18:32







user147263

















asked Oct 28 '15 at 14:22









Kees TilKees Til

602618




602618












  • $begingroup$
    Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
    $endgroup$
    – Jack D'Aurizio
    Oct 28 '15 at 14:25






  • 1




    $begingroup$
    1) makes no sense as written. Also such "question dumps" are not really welcome.
    $endgroup$
    – quid♦
    Oct 28 '15 at 14:30






  • 1




    $begingroup$
    What is $x$? Is it solve for $x$?
    $endgroup$
    – SchrodingersCat
    Oct 28 '15 at 14:33












  • $begingroup$
    sorry forgot the sentence has a solution i will edit it right away
    $endgroup$
    – Kees Til
    Oct 28 '15 at 14:36






  • 2




    $begingroup$
    The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
    $endgroup$
    – André Nicolas
    Oct 28 '15 at 16:38


















  • $begingroup$
    Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
    $endgroup$
    – Jack D'Aurizio
    Oct 28 '15 at 14:25






  • 1




    $begingroup$
    1) makes no sense as written. Also such "question dumps" are not really welcome.
    $endgroup$
    – quid♦
    Oct 28 '15 at 14:30






  • 1




    $begingroup$
    What is $x$? Is it solve for $x$?
    $endgroup$
    – SchrodingersCat
    Oct 28 '15 at 14:33












  • $begingroup$
    sorry forgot the sentence has a solution i will edit it right away
    $endgroup$
    – Kees Til
    Oct 28 '15 at 14:36






  • 2




    $begingroup$
    The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
    $endgroup$
    – André Nicolas
    Oct 28 '15 at 16:38
















$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25




$begingroup$
Have a look at this rather famous question: math.stackexchange.com/questions/685958/…
$endgroup$
– Jack D'Aurizio
Oct 28 '15 at 14:25




1




1




$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30




$begingroup$
1) makes no sense as written. Also such "question dumps" are not really welcome.
$endgroup$
– quid♦
Oct 28 '15 at 14:30




1




1




$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33






$begingroup$
What is $x$? Is it solve for $x$?
$endgroup$
– SchrodingersCat
Oct 28 '15 at 14:33














$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36




$begingroup$
sorry forgot the sentence has a solution i will edit it right away
$endgroup$
– Kees Til
Oct 28 '15 at 14:36




2




2




$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38




$begingroup$
The most straightforward way to evaluate most Legendre symbols $(a/p)$ is by using Quadratic Reciprocity. Is that one of the tools you already have?
$endgroup$
– André Nicolas
Oct 28 '15 at 16:38










1 Answer
1






active

oldest

votes


















0












$begingroup$


  1. You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.


  2. Agree with your own answer except for the typo:
    $$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$


  3. Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.


  4. If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$







share|cite|improve this answer











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    $begingroup$


    1. You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.


    2. Agree with your own answer except for the typo:
      $$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$


    3. Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.


    4. If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$







    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$


      1. You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.


      2. Agree with your own answer except for the typo:
        $$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$


      3. Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.


      4. If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$







      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$


        1. You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.


        2. Agree with your own answer except for the typo:
          $$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$


        3. Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.


        4. If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$







        share|cite|improve this answer











        $endgroup$




        1. You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p equiv 1 mod 3$ then there is an $x ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.


        2. Agree with your own answer except for the typo:
          $$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(frac{-3}{p})=1$


        3. Indeed $D=1^2-4cdot 1cdot 1 = -3$. Effectively 2. says that $sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=frac{-1 pm sqrt{-3}}{2} = frac{-1 pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.


        4. If $p equiv -1 mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(frac{-3}{p})=-1$








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        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 '15 at 17:06

























        answered Nov 28 '15 at 14:17









        Maestro13Maestro13

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