Partition on $Z_p$












2












$begingroup$


I'm trying to understand what is established in Proofs from the Book (from Eigner and Ziegler) concerning the representation of numbers as a sum of two squares.



Consider $mathbb{Z}_p$, $p$ an odd prime. At first, the authors affirm that is always possible to partition ${1,2, dots, p-1}$ in quadruples ${x, -x, overline{x}, - overline{x}}$, where $-x$ is the additive inverse of $x$ and $overline{x}$ is the multiplicative inverse of $x$. If two of these numbers are equal, the class reduces to a pair of elements. I proved the last affirmation.



My doubt is: the authors affirm that, if $p-1=4m+2$ we will have just one pair and the other elements will be quadruples and, if $p-1=4m$, there will be 2 pairs. According to the number of these pairs, they establish the number of solutions of $x^2 equiv 1 mod p$.



All my examples indicates that it is true (of course it is because the book is a classic!), but I can't understand it.






number-theory prime-numbers equivalence-relations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I'm trying to understand what is established in Proofs from the Book (from Eigner and Ziegler) concerning the representation of numbers as a sum of two squares.



    Consider $mathbb{Z}_p$, $p$ an odd prime. At first, the authors affirm that is always possible to partition ${1,2, dots, p-1}$ in quadruples ${x, -x, overline{x}, - overline{x}}$, where $-x$ is the additive inverse of $x$ and $overline{x}$ is the multiplicative inverse of $x$. If two of these numbers are equal, the class reduces to a pair of elements. I proved the last affirmation.



    My doubt is: the authors affirm that, if $p-1=4m+2$ we will have just one pair and the other elements will be quadruples and, if $p-1=4m$, there will be 2 pairs. According to the number of these pairs, they establish the number of solutions of $x^2 equiv 1 mod p$.



    All my examples indicates that it is true (of course it is because the book is a classic!), but I can't understand it.






    number-theory prime-numbers equivalence-relations







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I'm trying to understand what is established in Proofs from the Book (from Eigner and Ziegler) concerning the representation of numbers as a sum of two squares.



      Consider $mathbb{Z}_p$, $p$ an odd prime. At first, the authors affirm that is always possible to partition ${1,2, dots, p-1}$ in quadruples ${x, -x, overline{x}, - overline{x}}$, where $-x$ is the additive inverse of $x$ and $overline{x}$ is the multiplicative inverse of $x$. If two of these numbers are equal, the class reduces to a pair of elements. I proved the last affirmation.



      My doubt is: the authors affirm that, if $p-1=4m+2$ we will have just one pair and the other elements will be quadruples and, if $p-1=4m$, there will be 2 pairs. According to the number of these pairs, they establish the number of solutions of $x^2 equiv 1 mod p$.



      All my examples indicates that it is true (of course it is because the book is a classic!), but I can't understand it.






      number-theory prime-numbers equivalence-relations







      share|cite|improve this question











      $endgroup$




      I'm trying to understand what is established in Proofs from the Book (from Eigner and Ziegler) concerning the representation of numbers as a sum of two squares.



      Consider $mathbb{Z}_p$, $p$ an odd prime. At first, the authors affirm that is always possible to partition ${1,2, dots, p-1}$ in quadruples ${x, -x, overline{x}, - overline{x}}$, where $-x$ is the additive inverse of $x$ and $overline{x}$ is the multiplicative inverse of $x$. If two of these numbers are equal, the class reduces to a pair of elements. I proved the last affirmation.



      My doubt is: the authors affirm that, if $p-1=4m+2$ we will have just one pair and the other elements will be quadruples and, if $p-1=4m$, there will be 2 pairs. According to the number of these pairs, they establish the number of solutions of $x^2 equiv 1 mod p$.



      All my examples indicates that it is true (of course it is because the book is a classic!), but I can't understand it.






      number-theory prime-numbers equivalence-relations




      number-theory prime-numbers equivalence-relations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 30 at 20:18









      Jordan Green

      1,146410




      1,146410










      asked Jan 30 at 18:30









      Cleto PereiraCleto Pereira

      459




      459






















          1 Answer
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          $begingroup$

          If $p-1 = 4m+2$ for some $m$, then the number of pairs must be odd. (Otherwise, after taking the union of all quadruples and pairs, we’d get a total number of elements that was divisible by $4$.)
          Similarly, if $p−1=4m$, then the number of pairs must be even. Thus, to prove the authors’ claim, it suffices to prove that there are at most two pairs, regardless of the value of $p-1$ modulo $4$.



          The key to proving that there are at most two pairs is the fact that if $p$ is prime, then a degree $n$ polynomial can have at most $n$ roots modulo $p$. This can be proved via induction. (See, e.g. If $f$ is a polynomial of degree $n$, then $f(x) equiv 0pmod p$ has at most $n$ solutions.). Here, the relevant polynomials are $f(x) = x^2 -1$ and $g(x) = x^2+1$, because a quadruple ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ are both roots of $f$ or they are both roots of $g$.
          You probably proved something like this in the course of proving that if two elements of ${x,-x, overline{x}, - overline{x} }$ are equal, then it reduces to a pair; however, I’ve appended a proof for the sake of completeness.*



          Since $f$ and $g$ each have at most $2$ roots mod $p$, at most one quadruple can be reduced to a pair by way of solving each of $f$ and $g$. Thus, at most two quadruples can be reduced to pairs.



          We note that $f$ always has two solutions mod $p$, namely $1$ and $p-1$. Thus, the one pair we get when $p = 4m+1$ must result from roots of $f$, not roots of $g$. (The roots of $f$ and $g$ never coincide assuming that $p$ is odd.)




          *Appendix: Proof that ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ solve $f$ or $x$ and $-x$ solve $g$.



          By setting pairs from ${x,-x, overline{x}, - overline{x} }$ equal, one obtains six equations:
          $$
          x = -x tag{1};
          $$



          $$
          x = overline{x}; tag{2}
          $$



          $$
          x = - overline{x}; tag{3}
          $$



          $$
          -x = overline{x}; tag{4}
          $$



          $$
          -x = - overline{x}; tag{5}
          $$



          $$
          overline{x} = -overline{x}. tag{6}
          $$



          However, (1) and (6) are impossible for an odd $p$, while (2) is equivalent to (5) and (3) is equivalent to (4). Thus, the quadruple reduces to a pair if and only if (2) is satisfied or (3) is satisfied. In turn, (2) is satisfied if and only if $x$ and $-x$ are both roots of $f$, whereas (3) is satisfied if and only if $x$ and $-x$ are roots of $g$.






          share|cite|improve this answer











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            $begingroup$

            If $p-1 = 4m+2$ for some $m$, then the number of pairs must be odd. (Otherwise, after taking the union of all quadruples and pairs, we’d get a total number of elements that was divisible by $4$.)
            Similarly, if $p−1=4m$, then the number of pairs must be even. Thus, to prove the authors’ claim, it suffices to prove that there are at most two pairs, regardless of the value of $p-1$ modulo $4$.



            The key to proving that there are at most two pairs is the fact that if $p$ is prime, then a degree $n$ polynomial can have at most $n$ roots modulo $p$. This can be proved via induction. (See, e.g. If $f$ is a polynomial of degree $n$, then $f(x) equiv 0pmod p$ has at most $n$ solutions.). Here, the relevant polynomials are $f(x) = x^2 -1$ and $g(x) = x^2+1$, because a quadruple ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ are both roots of $f$ or they are both roots of $g$.
            You probably proved something like this in the course of proving that if two elements of ${x,-x, overline{x}, - overline{x} }$ are equal, then it reduces to a pair; however, I’ve appended a proof for the sake of completeness.*



            Since $f$ and $g$ each have at most $2$ roots mod $p$, at most one quadruple can be reduced to a pair by way of solving each of $f$ and $g$. Thus, at most two quadruples can be reduced to pairs.



            We note that $f$ always has two solutions mod $p$, namely $1$ and $p-1$. Thus, the one pair we get when $p = 4m+1$ must result from roots of $f$, not roots of $g$. (The roots of $f$ and $g$ never coincide assuming that $p$ is odd.)




            *Appendix: Proof that ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ solve $f$ or $x$ and $-x$ solve $g$.



            By setting pairs from ${x,-x, overline{x}, - overline{x} }$ equal, one obtains six equations:
            $$
            x = -x tag{1};
            $$



            $$
            x = overline{x}; tag{2}
            $$



            $$
            x = - overline{x}; tag{3}
            $$



            $$
            -x = overline{x}; tag{4}
            $$



            $$
            -x = - overline{x}; tag{5}
            $$



            $$
            overline{x} = -overline{x}. tag{6}
            $$



            However, (1) and (6) are impossible for an odd $p$, while (2) is equivalent to (5) and (3) is equivalent to (4). Thus, the quadruple reduces to a pair if and only if (2) is satisfied or (3) is satisfied. In turn, (2) is satisfied if and only if $x$ and $-x$ are both roots of $f$, whereas (3) is satisfied if and only if $x$ and $-x$ are roots of $g$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              If $p-1 = 4m+2$ for some $m$, then the number of pairs must be odd. (Otherwise, after taking the union of all quadruples and pairs, we’d get a total number of elements that was divisible by $4$.)
              Similarly, if $p−1=4m$, then the number of pairs must be even. Thus, to prove the authors’ claim, it suffices to prove that there are at most two pairs, regardless of the value of $p-1$ modulo $4$.



              The key to proving that there are at most two pairs is the fact that if $p$ is prime, then a degree $n$ polynomial can have at most $n$ roots modulo $p$. This can be proved via induction. (See, e.g. If $f$ is a polynomial of degree $n$, then $f(x) equiv 0pmod p$ has at most $n$ solutions.). Here, the relevant polynomials are $f(x) = x^2 -1$ and $g(x) = x^2+1$, because a quadruple ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ are both roots of $f$ or they are both roots of $g$.
              You probably proved something like this in the course of proving that if two elements of ${x,-x, overline{x}, - overline{x} }$ are equal, then it reduces to a pair; however, I’ve appended a proof for the sake of completeness.*



              Since $f$ and $g$ each have at most $2$ roots mod $p$, at most one quadruple can be reduced to a pair by way of solving each of $f$ and $g$. Thus, at most two quadruples can be reduced to pairs.



              We note that $f$ always has two solutions mod $p$, namely $1$ and $p-1$. Thus, the one pair we get when $p = 4m+1$ must result from roots of $f$, not roots of $g$. (The roots of $f$ and $g$ never coincide assuming that $p$ is odd.)




              *Appendix: Proof that ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ solve $f$ or $x$ and $-x$ solve $g$.



              By setting pairs from ${x,-x, overline{x}, - overline{x} }$ equal, one obtains six equations:
              $$
              x = -x tag{1};
              $$



              $$
              x = overline{x}; tag{2}
              $$



              $$
              x = - overline{x}; tag{3}
              $$



              $$
              -x = overline{x}; tag{4}
              $$



              $$
              -x = - overline{x}; tag{5}
              $$



              $$
              overline{x} = -overline{x}. tag{6}
              $$



              However, (1) and (6) are impossible for an odd $p$, while (2) is equivalent to (5) and (3) is equivalent to (4). Thus, the quadruple reduces to a pair if and only if (2) is satisfied or (3) is satisfied. In turn, (2) is satisfied if and only if $x$ and $-x$ are both roots of $f$, whereas (3) is satisfied if and only if $x$ and $-x$ are roots of $g$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                If $p-1 = 4m+2$ for some $m$, then the number of pairs must be odd. (Otherwise, after taking the union of all quadruples and pairs, we’d get a total number of elements that was divisible by $4$.)
                Similarly, if $p−1=4m$, then the number of pairs must be even. Thus, to prove the authors’ claim, it suffices to prove that there are at most two pairs, regardless of the value of $p-1$ modulo $4$.



                The key to proving that there are at most two pairs is the fact that if $p$ is prime, then a degree $n$ polynomial can have at most $n$ roots modulo $p$. This can be proved via induction. (See, e.g. If $f$ is a polynomial of degree $n$, then $f(x) equiv 0pmod p$ has at most $n$ solutions.). Here, the relevant polynomials are $f(x) = x^2 -1$ and $g(x) = x^2+1$, because a quadruple ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ are both roots of $f$ or they are both roots of $g$.
                You probably proved something like this in the course of proving that if two elements of ${x,-x, overline{x}, - overline{x} }$ are equal, then it reduces to a pair; however, I’ve appended a proof for the sake of completeness.*



                Since $f$ and $g$ each have at most $2$ roots mod $p$, at most one quadruple can be reduced to a pair by way of solving each of $f$ and $g$. Thus, at most two quadruples can be reduced to pairs.



                We note that $f$ always has two solutions mod $p$, namely $1$ and $p-1$. Thus, the one pair we get when $p = 4m+1$ must result from roots of $f$, not roots of $g$. (The roots of $f$ and $g$ never coincide assuming that $p$ is odd.)




                *Appendix: Proof that ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ solve $f$ or $x$ and $-x$ solve $g$.



                By setting pairs from ${x,-x, overline{x}, - overline{x} }$ equal, one obtains six equations:
                $$
                x = -x tag{1};
                $$



                $$
                x = overline{x}; tag{2}
                $$



                $$
                x = - overline{x}; tag{3}
                $$



                $$
                -x = overline{x}; tag{4}
                $$



                $$
                -x = - overline{x}; tag{5}
                $$



                $$
                overline{x} = -overline{x}. tag{6}
                $$



                However, (1) and (6) are impossible for an odd $p$, while (2) is equivalent to (5) and (3) is equivalent to (4). Thus, the quadruple reduces to a pair if and only if (2) is satisfied or (3) is satisfied. In turn, (2) is satisfied if and only if $x$ and $-x$ are both roots of $f$, whereas (3) is satisfied if and only if $x$ and $-x$ are roots of $g$.






                share|cite|improve this answer











                $endgroup$



                If $p-1 = 4m+2$ for some $m$, then the number of pairs must be odd. (Otherwise, after taking the union of all quadruples and pairs, we’d get a total number of elements that was divisible by $4$.)
                Similarly, if $p−1=4m$, then the number of pairs must be even. Thus, to prove the authors’ claim, it suffices to prove that there are at most two pairs, regardless of the value of $p-1$ modulo $4$.



                The key to proving that there are at most two pairs is the fact that if $p$ is prime, then a degree $n$ polynomial can have at most $n$ roots modulo $p$. This can be proved via induction. (See, e.g. If $f$ is a polynomial of degree $n$, then $f(x) equiv 0pmod p$ has at most $n$ solutions.). Here, the relevant polynomials are $f(x) = x^2 -1$ and $g(x) = x^2+1$, because a quadruple ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ are both roots of $f$ or they are both roots of $g$.
                You probably proved something like this in the course of proving that if two elements of ${x,-x, overline{x}, - overline{x} }$ are equal, then it reduces to a pair; however, I’ve appended a proof for the sake of completeness.*



                Since $f$ and $g$ each have at most $2$ roots mod $p$, at most one quadruple can be reduced to a pair by way of solving each of $f$ and $g$. Thus, at most two quadruples can be reduced to pairs.



                We note that $f$ always has two solutions mod $p$, namely $1$ and $p-1$. Thus, the one pair we get when $p = 4m+1$ must result from roots of $f$, not roots of $g$. (The roots of $f$ and $g$ never coincide assuming that $p$ is odd.)




                *Appendix: Proof that ${x,-x, overline{x}, - overline{x} }$ reduces to a pair if and only if $x$ and $-x$ solve $f$ or $x$ and $-x$ solve $g$.



                By setting pairs from ${x,-x, overline{x}, - overline{x} }$ equal, one obtains six equations:
                $$
                x = -x tag{1};
                $$



                $$
                x = overline{x}; tag{2}
                $$



                $$
                x = - overline{x}; tag{3}
                $$



                $$
                -x = overline{x}; tag{4}
                $$



                $$
                -x = - overline{x}; tag{5}
                $$



                $$
                overline{x} = -overline{x}. tag{6}
                $$



                However, (1) and (6) are impossible for an odd $p$, while (2) is equivalent to (5) and (3) is equivalent to (4). Thus, the quadruple reduces to a pair if and only if (2) is satisfied or (3) is satisfied. In turn, (2) is satisfied if and only if $x$ and $-x$ are both roots of $f$, whereas (3) is satisfied if and only if $x$ and $-x$ are roots of $g$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 30 at 20:11

























                answered Jan 30 at 19:55









                Jordan GreenJordan Green

                1,146410




                1,146410






























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