What primes $p$ give solutions to $x^{2} equiv 7 ($mod $ p)$











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I'm trying to understand how to solve this using the Legendre symbol but am having a hard time figuring out exactly what to do.



There are many different cases to consider but I do not know how to approach this problem.










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  • What are you stuck on? Do you know what the Legendre symbol actually is?
    – anomaly
    Nov 28 at 0:58












  • It tells us whether we have a quadratic or non quadratic residue. I'm just not too familiar with computing it
    – Wallace
    Nov 28 at 1:01















up vote
-1
down vote

favorite












I'm trying to understand how to solve this using the Legendre symbol but am having a hard time figuring out exactly what to do.



There are many different cases to consider but I do not know how to approach this problem.










share|cite|improve this question
























  • What are you stuck on? Do you know what the Legendre symbol actually is?
    – anomaly
    Nov 28 at 0:58












  • It tells us whether we have a quadratic or non quadratic residue. I'm just not too familiar with computing it
    – Wallace
    Nov 28 at 1:01













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I'm trying to understand how to solve this using the Legendre symbol but am having a hard time figuring out exactly what to do.



There are many different cases to consider but I do not know how to approach this problem.










share|cite|improve this question















I'm trying to understand how to solve this using the Legendre symbol but am having a hard time figuring out exactly what to do.



There are many different cases to consider but I do not know how to approach this problem.







number-theory quadratic-residues






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edited Nov 28 at 1:08









Henning Makholm

236k16300534




236k16300534










asked Nov 28 at 0:38









Wallace

1226




1226












  • What are you stuck on? Do you know what the Legendre symbol actually is?
    – anomaly
    Nov 28 at 0:58












  • It tells us whether we have a quadratic or non quadratic residue. I'm just not too familiar with computing it
    – Wallace
    Nov 28 at 1:01


















  • What are you stuck on? Do you know what the Legendre symbol actually is?
    – anomaly
    Nov 28 at 0:58












  • It tells us whether we have a quadratic or non quadratic residue. I'm just not too familiar with computing it
    – Wallace
    Nov 28 at 1:01
















What are you stuck on? Do you know what the Legendre symbol actually is?
– anomaly
Nov 28 at 0:58






What are you stuck on? Do you know what the Legendre symbol actually is?
– anomaly
Nov 28 at 0:58














It tells us whether we have a quadratic or non quadratic residue. I'm just not too familiar with computing it
– Wallace
Nov 28 at 1:01




It tells us whether we have a quadratic or non quadratic residue. I'm just not too familiar with computing it
– Wallace
Nov 28 at 1:01










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










Another hint:



An odd prime is congruent to $1$ or $3mod 4$. Also, the non-zero squares modulo $7$ are $;1,,2$ and $4$. So




  • If $pequiv 1mod 4$, $(-1)^{tfrac{p-1}2}=1$, and $p$ has to be a square modulo $7$, i.e. $pequiv 1,2,4$.

  • If $pequiv 3mod 4$, $p$ has to be a non-square modulo $7$, i.e. $pequiv3,5,6$


Now use the Chinese remainder theorem:
$$mathbf Z/4mathbf Ztimes mathbf Z/7mathbf Zsimeqmathbf Z/28mathbf Z$$
to find the images of the pairs ${(1,1),,(1,2),,(1,4),,(3,3),,(3,5),,(3,6)}$ modulo $28$.






share|cite|improve this answer






























    up vote
    2
    down vote













    Hint: If $p$ is an odd prime not equal to $7$ ($p=2$ works, $p=7$ does not), we have by quadratic reciprocity that



    $$left(frac{p}{7}right)left(frac{7}{p}right)=(-1)^{frac{6(p-1)}{4}}=(-1)^{frac{p-1}{2}},$$



    so we need



    $$left(frac{p}{7}right)=(-1)^{frac{p-1}{2}}.$$



    Now look at each residue for $pbmod 7$ and determine what residue $pbmod 4$ must be.






    share|cite|improve this answer




























      up vote
      0
      down vote













      If either $p$ or $-p$ has an integer expression as $u^2 - 7 v^2,$ it follows that there is a solution to $$ u^2 equiv 7 v^2 pmod p ; . ; $$
      Which, you know, means something.



           2    29    37    53   109   113   137   149   193   197
      233 277 281 317 337 373 389 401 421 449
      457 541 557 569 613 617 641 653 673 701
      709 757 809 821 877 953 977 1009 1033 1061

      3 7 19 31 47 59 83 103 131 139
      167 199 223 227 251 271 283 307 311 367
      383 419 439 467 479 503 523 563 587 607
      619 643 647 691 719 727 787 811 839 859
      887 971 983 1039 1063 1091 1123 1151 1223 1231





      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        Another hint:



        An odd prime is congruent to $1$ or $3mod 4$. Also, the non-zero squares modulo $7$ are $;1,,2$ and $4$. So




        • If $pequiv 1mod 4$, $(-1)^{tfrac{p-1}2}=1$, and $p$ has to be a square modulo $7$, i.e. $pequiv 1,2,4$.

        • If $pequiv 3mod 4$, $p$ has to be a non-square modulo $7$, i.e. $pequiv3,5,6$


        Now use the Chinese remainder theorem:
        $$mathbf Z/4mathbf Ztimes mathbf Z/7mathbf Zsimeqmathbf Z/28mathbf Z$$
        to find the images of the pairs ${(1,1),,(1,2),,(1,4),,(3,3),,(3,5),,(3,6)}$ modulo $28$.






        share|cite|improve this answer



























          up vote
          2
          down vote



          accepted










          Another hint:



          An odd prime is congruent to $1$ or $3mod 4$. Also, the non-zero squares modulo $7$ are $;1,,2$ and $4$. So




          • If $pequiv 1mod 4$, $(-1)^{tfrac{p-1}2}=1$, and $p$ has to be a square modulo $7$, i.e. $pequiv 1,2,4$.

          • If $pequiv 3mod 4$, $p$ has to be a non-square modulo $7$, i.e. $pequiv3,5,6$


          Now use the Chinese remainder theorem:
          $$mathbf Z/4mathbf Ztimes mathbf Z/7mathbf Zsimeqmathbf Z/28mathbf Z$$
          to find the images of the pairs ${(1,1),,(1,2),,(1,4),,(3,3),,(3,5),,(3,6)}$ modulo $28$.






          share|cite|improve this answer

























            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Another hint:



            An odd prime is congruent to $1$ or $3mod 4$. Also, the non-zero squares modulo $7$ are $;1,,2$ and $4$. So




            • If $pequiv 1mod 4$, $(-1)^{tfrac{p-1}2}=1$, and $p$ has to be a square modulo $7$, i.e. $pequiv 1,2,4$.

            • If $pequiv 3mod 4$, $p$ has to be a non-square modulo $7$, i.e. $pequiv3,5,6$


            Now use the Chinese remainder theorem:
            $$mathbf Z/4mathbf Ztimes mathbf Z/7mathbf Zsimeqmathbf Z/28mathbf Z$$
            to find the images of the pairs ${(1,1),,(1,2),,(1,4),,(3,3),,(3,5),,(3,6)}$ modulo $28$.






            share|cite|improve this answer














            Another hint:



            An odd prime is congruent to $1$ or $3mod 4$. Also, the non-zero squares modulo $7$ are $;1,,2$ and $4$. So




            • If $pequiv 1mod 4$, $(-1)^{tfrac{p-1}2}=1$, and $p$ has to be a square modulo $7$, i.e. $pequiv 1,2,4$.

            • If $pequiv 3mod 4$, $p$ has to be a non-square modulo $7$, i.e. $pequiv3,5,6$


            Now use the Chinese remainder theorem:
            $$mathbf Z/4mathbf Ztimes mathbf Z/7mathbf Zsimeqmathbf Z/28mathbf Z$$
            to find the images of the pairs ${(1,1),,(1,2),,(1,4),,(3,3),,(3,5),,(3,6)}$ modulo $28$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 28 at 9:25

























            answered Nov 28 at 1:06









            Bernard

            117k637109




            117k637109






















                up vote
                2
                down vote













                Hint: If $p$ is an odd prime not equal to $7$ ($p=2$ works, $p=7$ does not), we have by quadratic reciprocity that



                $$left(frac{p}{7}right)left(frac{7}{p}right)=(-1)^{frac{6(p-1)}{4}}=(-1)^{frac{p-1}{2}},$$



                so we need



                $$left(frac{p}{7}right)=(-1)^{frac{p-1}{2}}.$$



                Now look at each residue for $pbmod 7$ and determine what residue $pbmod 4$ must be.






                share|cite|improve this answer

























                  up vote
                  2
                  down vote













                  Hint: If $p$ is an odd prime not equal to $7$ ($p=2$ works, $p=7$ does not), we have by quadratic reciprocity that



                  $$left(frac{p}{7}right)left(frac{7}{p}right)=(-1)^{frac{6(p-1)}{4}}=(-1)^{frac{p-1}{2}},$$



                  so we need



                  $$left(frac{p}{7}right)=(-1)^{frac{p-1}{2}}.$$



                  Now look at each residue for $pbmod 7$ and determine what residue $pbmod 4$ must be.






                  share|cite|improve this answer























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Hint: If $p$ is an odd prime not equal to $7$ ($p=2$ works, $p=7$ does not), we have by quadratic reciprocity that



                    $$left(frac{p}{7}right)left(frac{7}{p}right)=(-1)^{frac{6(p-1)}{4}}=(-1)^{frac{p-1}{2}},$$



                    so we need



                    $$left(frac{p}{7}right)=(-1)^{frac{p-1}{2}}.$$



                    Now look at each residue for $pbmod 7$ and determine what residue $pbmod 4$ must be.






                    share|cite|improve this answer












                    Hint: If $p$ is an odd prime not equal to $7$ ($p=2$ works, $p=7$ does not), we have by quadratic reciprocity that



                    $$left(frac{p}{7}right)left(frac{7}{p}right)=(-1)^{frac{6(p-1)}{4}}=(-1)^{frac{p-1}{2}},$$



                    so we need



                    $$left(frac{p}{7}right)=(-1)^{frac{p-1}{2}}.$$



                    Now look at each residue for $pbmod 7$ and determine what residue $pbmod 4$ must be.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 28 at 0:51









                    Carl Schildkraut

                    11k11439




                    11k11439






















                        up vote
                        0
                        down vote













                        If either $p$ or $-p$ has an integer expression as $u^2 - 7 v^2,$ it follows that there is a solution to $$ u^2 equiv 7 v^2 pmod p ; . ; $$
                        Which, you know, means something.



                             2    29    37    53   109   113   137   149   193   197
                        233 277 281 317 337 373 389 401 421 449
                        457 541 557 569 613 617 641 653 673 701
                        709 757 809 821 877 953 977 1009 1033 1061

                        3 7 19 31 47 59 83 103 131 139
                        167 199 223 227 251 271 283 307 311 367
                        383 419 439 467 479 503 523 563 587 607
                        619 643 647 691 719 727 787 811 839 859
                        887 971 983 1039 1063 1091 1123 1151 1223 1231





                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          If either $p$ or $-p$ has an integer expression as $u^2 - 7 v^2,$ it follows that there is a solution to $$ u^2 equiv 7 v^2 pmod p ; . ; $$
                          Which, you know, means something.



                               2    29    37    53   109   113   137   149   193   197
                          233 277 281 317 337 373 389 401 421 449
                          457 541 557 569 613 617 641 653 673 701
                          709 757 809 821 877 953 977 1009 1033 1061

                          3 7 19 31 47 59 83 103 131 139
                          167 199 223 227 251 271 283 307 311 367
                          383 419 439 467 479 503 523 563 587 607
                          619 643 647 691 719 727 787 811 839 859
                          887 971 983 1039 1063 1091 1123 1151 1223 1231





                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            If either $p$ or $-p$ has an integer expression as $u^2 - 7 v^2,$ it follows that there is a solution to $$ u^2 equiv 7 v^2 pmod p ; . ; $$
                            Which, you know, means something.



                                 2    29    37    53   109   113   137   149   193   197
                            233 277 281 317 337 373 389 401 421 449
                            457 541 557 569 613 617 641 653 673 701
                            709 757 809 821 877 953 977 1009 1033 1061

                            3 7 19 31 47 59 83 103 131 139
                            167 199 223 227 251 271 283 307 311 367
                            383 419 439 467 479 503 523 563 587 607
                            619 643 647 691 719 727 787 811 839 859
                            887 971 983 1039 1063 1091 1123 1151 1223 1231





                            share|cite|improve this answer












                            If either $p$ or $-p$ has an integer expression as $u^2 - 7 v^2,$ it follows that there is a solution to $$ u^2 equiv 7 v^2 pmod p ; . ; $$
                            Which, you know, means something.



                                 2    29    37    53   109   113   137   149   193   197
                            233 277 281 317 337 373 389 401 421 449
                            457 541 557 569 613 617 641 653 673 701
                            709 757 809 821 877 953 977 1009 1033 1061

                            3 7 19 31 47 59 83 103 131 139
                            167 199 223 227 251 271 283 307 311 367
                            383 419 439 467 479 503 523 563 587 607
                            619 643 647 691 719 727 787 811 839 859
                            887 971 983 1039 1063 1091 1123 1151 1223 1231






                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 28 at 2:28









                            Will Jagy

                            101k598198




                            101k598198






























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