Are rational points dense on every circle in the coordinate plane?

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Are rational points dense on every circle in the coordinate plane?



First thing first I know that rational points are dense on the unit circle.
However, I am not so sure how to show that rational points are not dense on every circle.



How would one come about answering this.
Any hits are appreciate it.






coordinate-systems rational-numbers







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    up vote
    1
    down vote

    favorite












    Are rational points dense on every circle in the coordinate plane?



    First thing first I know that rational points are dense on the unit circle.
    However, I am not so sure how to show that rational points are not dense on every circle.



    How would one come about answering this.
    Any hits are appreciate it.






    coordinate-systems rational-numbers







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Are rational points dense on every circle in the coordinate plane?



      First thing first I know that rational points are dense on the unit circle.
      However, I am not so sure how to show that rational points are not dense on every circle.



      How would one come about answering this.
      Any hits are appreciate it.






      coordinate-systems rational-numbers







      share|cite|improve this question















      Are rational points dense on every circle in the coordinate plane?



      First thing first I know that rational points are dense on the unit circle.
      However, I am not so sure how to show that rational points are not dense on every circle.



      How would one come about answering this.
      Any hits are appreciate it.






      coordinate-systems rational-numbers




      coordinate-systems rational-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago

























      asked 2 hours ago









      Hidaw

      110219




      110219




















          2 Answers
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          In particular, $x^2+y^2=3$ cannot have any rational points. If it had any such points then there would be integers $a,b,c$ having no common factor such that $(a/c)^2+(b/c)^2=3$ therefore $a^2+b^2=3c^2$. But with no common factor at least one of $a,b,c$ must be odd and all possibilities conforming with this requirement fail $bmod 4$.






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            up vote
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            They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles, so most circles contain no rational points at all.






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            • 1




              For example, the circle with center at the origin and radius $pi$ has no rational points. Replace $pi$ with any number whose square is irrational.
              – GEdgar
              2 hours ago











            • I had to read this six times before it hit me. Amazing.
              – Randall
              2 hours ago










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            2 Answers
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            2 Answers
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            up vote
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            In particular, $x^2+y^2=3$ cannot have any rational points. If it had any such points then there would be integers $a,b,c$ having no common factor such that $(a/c)^2+(b/c)^2=3$ therefore $a^2+b^2=3c^2$. But with no common factor at least one of $a,b,c$ must be odd and all possibilities conforming with this requirement fail $bmod 4$.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              In particular, $x^2+y^2=3$ cannot have any rational points. If it had any such points then there would be integers $a,b,c$ having no common factor such that $(a/c)^2+(b/c)^2=3$ therefore $a^2+b^2=3c^2$. But with no common factor at least one of $a,b,c$ must be odd and all possibilities conforming with this requirement fail $bmod 4$.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                In particular, $x^2+y^2=3$ cannot have any rational points. If it had any such points then there would be integers $a,b,c$ having no common factor such that $(a/c)^2+(b/c)^2=3$ therefore $a^2+b^2=3c^2$. But with no common factor at least one of $a,b,c$ must be odd and all possibilities conforming with this requirement fail $bmod 4$.






                share|cite|improve this answer












                In particular, $x^2+y^2=3$ cannot have any rational points. If it had any such points then there would be integers $a,b,c$ having no common factor such that $(a/c)^2+(b/c)^2=3$ therefore $a^2+b^2=3c^2$. But with no common factor at least one of $a,b,c$ must be odd and all possibilities conforming with this requirement fail $bmod 4$.







                share|cite|improve this answer












                share|cite|improve this answer



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                answered 38 mins ago









                Oscar Lanzi

                10.8k11734




                10.8k11734




















                    up vote
                    5
                    down vote













                    They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles, so most circles contain no rational points at all.






                    share|cite|improve this answer
















                    • 1




                      For example, the circle with center at the origin and radius $pi$ has no rational points. Replace $pi$ with any number whose square is irrational.
                      – GEdgar
                      2 hours ago











                    • I had to read this six times before it hit me. Amazing.
                      – Randall
                      2 hours ago














                    up vote
                    5
                    down vote













                    They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles, so most circles contain no rational points at all.






                    share|cite|improve this answer
















                    • 1




                      For example, the circle with center at the origin and radius $pi$ has no rational points. Replace $pi$ with any number whose square is irrational.
                      – GEdgar
                      2 hours ago











                    • I had to read this six times before it hit me. Amazing.
                      – Randall
                      2 hours ago












                    up vote
                    5
                    down vote










                    up vote
                    5
                    down vote









                    They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles, so most circles contain no rational points at all.






                    share|cite|improve this answer












                    They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles, so most circles contain no rational points at all.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    Matt Samuel

                    35.2k43060




                    35.2k43060







                    • 1




                      For example, the circle with center at the origin and radius $pi$ has no rational points. Replace $pi$ with any number whose square is irrational.
                      – GEdgar
                      2 hours ago











                    • I had to read this six times before it hit me. Amazing.
                      – Randall
                      2 hours ago












                    • 1




                      For example, the circle with center at the origin and radius $pi$ has no rational points. Replace $pi$ with any number whose square is irrational.
                      – GEdgar
                      2 hours ago











                    • I had to read this six times before it hit me. Amazing.
                      – Randall
                      2 hours ago







                    1




                    1




                    For example, the circle with center at the origin and radius $pi$ has no rational points. Replace $pi$ with any number whose square is irrational.
                    – GEdgar
                    2 hours ago





                    For example, the circle with center at the origin and radius $pi$ has no rational points. Replace $pi$ with any number whose square is irrational.
                    – GEdgar
                    2 hours ago













                    I had to read this six times before it hit me. Amazing.
                    – Randall
                    2 hours ago




                    I had to read this six times before it hit me. Amazing.
                    – Randall
                    2 hours ago

















                     

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