Primes dividing $x^2+xy+y^2+1$












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Is there infinitely prime numbers $p$ such that $p$ divides $x^2+xy+y^2+1$ for some integers $x,y in Bbb Z$ ?





I can show that every prime $p$ divides $x^2+y^2+1$ for some integers $x,y in Bbb Z$, see Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p|(x^{2} + y^{2} + 1) $. :
the sets ${-1-x^2 mid x in Bbb F_p}$ and ${y^2 mid y in Bbb F_p}$ have both cardinality $(p+1)/2$, so they must have a non-empty intersection, i.e. an element of the form $z=-1-x^2=y^2$.



But the proof does not work to show that all but finitely many (or at least infinitely many) primes divide $x^2+xy+y^2+1$ for some integers $x,y$. So what can I do?






number-theory polynomials prime-numbers







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    1
















    Is there infinitely prime numbers $p$ such that $p$ divides $x^2+xy+y^2+1$ for some integers $x,y in Bbb Z$ ?





    I can show that every prime $p$ divides $x^2+y^2+1$ for some integers $x,y in Bbb Z$, see Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p|(x^{2} + y^{2} + 1) $. :
    the sets ${-1-x^2 mid x in Bbb F_p}$ and ${y^2 mid y in Bbb F_p}$ have both cardinality $(p+1)/2$, so they must have a non-empty intersection, i.e. an element of the form $z=-1-x^2=y^2$.



    But the proof does not work to show that all but finitely many (or at least infinitely many) primes divide $x^2+xy+y^2+1$ for some integers $x,y$. So what can I do?






    number-theory polynomials prime-numbers







    share|cite|improve this question



























      1












      1








      1


      1







      Is there infinitely prime numbers $p$ such that $p$ divides $x^2+xy+y^2+1$ for some integers $x,y in Bbb Z$ ?





      I can show that every prime $p$ divides $x^2+y^2+1$ for some integers $x,y in Bbb Z$, see Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p|(x^{2} + y^{2} + 1) $. :
      the sets ${-1-x^2 mid x in Bbb F_p}$ and ${y^2 mid y in Bbb F_p}$ have both cardinality $(p+1)/2$, so they must have a non-empty intersection, i.e. an element of the form $z=-1-x^2=y^2$.



      But the proof does not work to show that all but finitely many (or at least infinitely many) primes divide $x^2+xy+y^2+1$ for some integers $x,y$. So what can I do?






      number-theory polynomials prime-numbers







      share|cite|improve this question

















      Is there infinitely prime numbers $p$ such that $p$ divides $x^2+xy+y^2+1$ for some integers $x,y in Bbb Z$ ?





      I can show that every prime $p$ divides $x^2+y^2+1$ for some integers $x,y in Bbb Z$, see Show that for any prime $ p $, there are integers $ x $ and $ y $ such that $ p|(x^{2} + y^{2} + 1) $. :
      the sets ${-1-x^2 mid x in Bbb F_p}$ and ${y^2 mid y in Bbb F_p}$ have both cardinality $(p+1)/2$, so they must have a non-empty intersection, i.e. an element of the form $z=-1-x^2=y^2$.



      But the proof does not work to show that all but finitely many (or at least infinitely many) primes divide $x^2+xy+y^2+1$ for some integers $x,y$. So what can I do?






      number-theory polynomials prime-numbers




      number-theory polynomials prime-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 at 8:26

























      asked Dec 2 at 11:27









      Alphonse

      2,178623




      2,178623






















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          If $pne2$ then $pmid(x^2+xy+y^2+1)$ iff $pmid(4x^2+4xy+4y^2+4)$
          that is iff $pmid((2x+y)^2+3y^2+4)$. The same trick as before proves that there
          are $y$ and $z$ with $pmid(z^2+3y^2+4)$.






          share|cite|improve this answer





















          • Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$...
            – Alphonse
            Dec 2 at 12:27








          • 1




            @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)equiv0$.
            – Lord Shark the Unknown
            Dec 2 at 12:33











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          If $pne2$ then $pmid(x^2+xy+y^2+1)$ iff $pmid(4x^2+4xy+4y^2+4)$
          that is iff $pmid((2x+y)^2+3y^2+4)$. The same trick as before proves that there
          are $y$ and $z$ with $pmid(z^2+3y^2+4)$.






          share|cite|improve this answer





















          • Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$...
            – Alphonse
            Dec 2 at 12:27








          • 1




            @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)equiv0$.
            – Lord Shark the Unknown
            Dec 2 at 12:33
















          1














          If $pne2$ then $pmid(x^2+xy+y^2+1)$ iff $pmid(4x^2+4xy+4y^2+4)$
          that is iff $pmid((2x+y)^2+3y^2+4)$. The same trick as before proves that there
          are $y$ and $z$ with $pmid(z^2+3y^2+4)$.






          share|cite|improve this answer





















          • Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$...
            – Alphonse
            Dec 2 at 12:27








          • 1




            @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)equiv0$.
            – Lord Shark the Unknown
            Dec 2 at 12:33














          1












          1








          1






          If $pne2$ then $pmid(x^2+xy+y^2+1)$ iff $pmid(4x^2+4xy+4y^2+4)$
          that is iff $pmid((2x+y)^2+3y^2+4)$. The same trick as before proves that there
          are $y$ and $z$ with $pmid(z^2+3y^2+4)$.






          share|cite|improve this answer












          If $pne2$ then $pmid(x^2+xy+y^2+1)$ iff $pmid(4x^2+4xy+4y^2+4)$
          that is iff $pmid((2x+y)^2+3y^2+4)$. The same trick as before proves that there
          are $y$ and $z$ with $pmid(z^2+3y^2+4)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 at 11:32









          Lord Shark the Unknown

          101k958132




          101k958132












          • Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$...
            – Alphonse
            Dec 2 at 12:27








          • 1




            @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)equiv0$.
            – Lord Shark the Unknown
            Dec 2 at 12:33


















          • Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$...
            – Alphonse
            Dec 2 at 12:27








          • 1




            @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)equiv0$.
            – Lord Shark the Unknown
            Dec 2 at 12:33
















          Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$...
          – Alphonse
          Dec 2 at 12:27






          Thank you. I was wondering if it was also possible to solve the case of primes dividing $x^2+5xy+y^2+1$, for instance. I shouldn't have fixed the coefficient of $xy$ to $1$ (so I'm asking for $5xy$, also). I was not able to generalize your method to deal with $x^2+5xy+y^2+1$, or $x^2+7xy+y^2+1$...
          – Alphonse
          Dec 2 at 12:27






          1




          1




          @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)equiv0$.
          – Lord Shark the Unknown
          Dec 2 at 12:33




          @Alphonse Yes, it's exactly the same trick. Anyway, for any non-constant quadratic $f$ in $x$ and $y$, there are modulo $p$ solutions to $f(x,y)equiv0$.
          – Lord Shark the Unknown
          Dec 2 at 12:33


















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