on the quadratic form $5x^2-y^2$












2












$begingroup$


Consider the following subsets of $mathbb{Z}$:



$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and



$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.



Are there elements in $B$ which are not in $A$?






number-theory quadratic-forms







share|cite|improve this question









$endgroup$












  • $begingroup$
    there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
    $endgroup$
    – Dog_69
    Jan 20 at 19:35










  • $begingroup$
    But you haven't prove that $1notin A$, have you? @Dog_69
    $endgroup$
    – greedoid
    Jan 20 at 19:36












  • $begingroup$
    No, I haven't. I'm pretty sure but I don't have any proof.
    $endgroup$
    – Dog_69
    Jan 20 at 19:40






  • 3




    $begingroup$
    Taking $p=5$ and $y=11$ shows that $1in A$.
    $endgroup$
    – user84909
    Jan 20 at 19:41






  • 1




    $begingroup$
    What is a minumum number in B that you can't prove it is in A?
    $endgroup$
    – greedoid
    Jan 20 at 19:42
















2












$begingroup$


Consider the following subsets of $mathbb{Z}$:



$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and



$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.



Are there elements in $B$ which are not in $A$?






number-theory quadratic-forms







share|cite|improve this question









$endgroup$












  • $begingroup$
    there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
    $endgroup$
    – Dog_69
    Jan 20 at 19:35










  • $begingroup$
    But you haven't prove that $1notin A$, have you? @Dog_69
    $endgroup$
    – greedoid
    Jan 20 at 19:36












  • $begingroup$
    No, I haven't. I'm pretty sure but I don't have any proof.
    $endgroup$
    – Dog_69
    Jan 20 at 19:40






  • 3




    $begingroup$
    Taking $p=5$ and $y=11$ shows that $1in A$.
    $endgroup$
    – user84909
    Jan 20 at 19:41






  • 1




    $begingroup$
    What is a minumum number in B that you can't prove it is in A?
    $endgroup$
    – greedoid
    Jan 20 at 19:42














2












2








2


0



$begingroup$


Consider the following subsets of $mathbb{Z}$:



$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and



$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.



Are there elements in $B$ which are not in $A$?






number-theory quadratic-forms







share|cite|improve this question









$endgroup$




Consider the following subsets of $mathbb{Z}$:



$A={frac{5p^2-y^2}{4}, | , p, y ,,text{odd positive integers}, p,, text{prime}}$ and



$B={frac{5x^2-y^2}{4}, | , x, y ,,text{odd positive integers}}$.



Are there elements in $B$ which are not in $A$?






number-theory quadratic-forms




number-theory quadratic-forms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 20 at 19:27









user84909user84909

1908




1908












  • $begingroup$
    there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
    $endgroup$
    – Dog_69
    Jan 20 at 19:35










  • $begingroup$
    But you haven't prove that $1notin A$, have you? @Dog_69
    $endgroup$
    – greedoid
    Jan 20 at 19:36












  • $begingroup$
    No, I haven't. I'm pretty sure but I don't have any proof.
    $endgroup$
    – Dog_69
    Jan 20 at 19:40






  • 3




    $begingroup$
    Taking $p=5$ and $y=11$ shows that $1in A$.
    $endgroup$
    – user84909
    Jan 20 at 19:41






  • 1




    $begingroup$
    What is a minumum number in B that you can't prove it is in A?
    $endgroup$
    – greedoid
    Jan 20 at 19:42


















  • $begingroup$
    there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
    $endgroup$
    – Dog_69
    Jan 20 at 19:35










  • $begingroup$
    But you haven't prove that $1notin A$, have you? @Dog_69
    $endgroup$
    – greedoid
    Jan 20 at 19:36












  • $begingroup$
    No, I haven't. I'm pretty sure but I don't have any proof.
    $endgroup$
    – Dog_69
    Jan 20 at 19:40






  • 3




    $begingroup$
    Taking $p=5$ and $y=11$ shows that $1in A$.
    $endgroup$
    – user84909
    Jan 20 at 19:41






  • 1




    $begingroup$
    What is a minumum number in B that you can't prove it is in A?
    $endgroup$
    – greedoid
    Jan 20 at 19:42
















$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35




$begingroup$
there's a problem with $1$ because it is not prime. Hence $1in B$ but I think $1notin A$. Apart from that very good question.
$endgroup$
– Dog_69
Jan 20 at 19:35












$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36






$begingroup$
But you haven't prove that $1notin A$, have you? @Dog_69
$endgroup$
– greedoid
Jan 20 at 19:36














$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40




$begingroup$
No, I haven't. I'm pretty sure but I don't have any proof.
$endgroup$
– Dog_69
Jan 20 at 19:40




3




3




$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41




$begingroup$
Taking $p=5$ and $y=11$ shows that $1in A$.
$endgroup$
– user84909
Jan 20 at 19:41




1




1




$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42




$begingroup$
What is a minumum number in B that you can't prove it is in A?
$endgroup$
– greedoid
Jan 20 at 19:42










1 Answer
1






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1












$begingroup$

Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396equiv0pmod{3},$$
from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99in B$ but $99notin A$.






share|cite|improve this answer









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    $begingroup$

    Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
    $$5p^2-y^2=396equiv0pmod{3},$$
    from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
    $$y^2=5p^2-396=-351,$$
    a contradiction. So $99in B$ but $99notin A$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
      $$5p^2-y^2=396equiv0pmod{3},$$
      from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
      $$y^2=5p^2-396=-351,$$
      a contradiction. So $99in B$ but $99notin A$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
        $$5p^2-y^2=396equiv0pmod{3},$$
        from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
        $$y^2=5p^2-396=-351,$$
        a contradiction. So $99in B$ but $99notin A$.






        share|cite|improve this answer









        $endgroup$



        Note that $frac{5cdot9^2-3^2}{4}=99$, and if $frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
        $$5p^2-y^2=396equiv0pmod{3},$$
        from which it follows that $pequiv yequiv0pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
        $$y^2=5p^2-396=-351,$$
        a contradiction. So $99in B$ but $99notin A$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 21:15









        ServaesServaes

        25.7k33996




        25.7k33996






























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