Prove that there are infinitely many primes $p$ such that $x^10 + x + 1 equiv 0 mod p$ has at least one solution $xinmathbbZ$.

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Prove that there are infinitely many primes $p$ such that



$$x^10 + x + 1 equiv 0 mod p$$



has at least one solution $xinmathbbZ$.



I believe I should be doing a proof by contradiction but I cannot figure out where it arises. Any help will be appreciated! thank you!






number-theory







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    up vote
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    favorite
    1












    Prove that there are infinitely many primes $p$ such that



    $$x^10 + x + 1 equiv 0 mod p$$



    has at least one solution $xinmathbbZ$.



    I believe I should be doing a proof by contradiction but I cannot figure out where it arises. Any help will be appreciated! thank you!






    number-theory







    share|cite|improve this question























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1





      Prove that there are infinitely many primes $p$ such that



      $$x^10 + x + 1 equiv 0 mod p$$



      has at least one solution $xinmathbbZ$.



      I believe I should be doing a proof by contradiction but I cannot figure out where it arises. Any help will be appreciated! thank you!






      number-theory







      share|cite|improve this question













      Prove that there are infinitely many primes $p$ such that



      $$x^10 + x + 1 equiv 0 mod p$$



      has at least one solution $xinmathbbZ$.



      I believe I should be doing a proof by contradiction but I cannot figure out where it arises. Any help will be appreciated! thank you!






      number-theory




      number-theory






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      asked Aug 30 at 7:08









      David

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          Yes, a proof by contradiction is a good way to proceed. Assume that the property holds only for a finite set of primes $p_1,p_2,dots, p_n$ then let
          $x:=p_1p_2cdots p_n-1$. It follows that for $i=1,dots,n$,
          $$N:=x^10+x+1equiv 1-1+1=1 pmodp_i.$$
          Now consider $q$ be a prime which divides $N$. What may we conclude?






          share|cite|improve this answer






















          • We can also take any finite set $S$ of primes such that $3in S$ and let $x=prod_pin Sp.$.............+1
            – DanielWainfleet
            Aug 30 at 15:31










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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          up vote
          9
          down vote



          accepted










          Yes, a proof by contradiction is a good way to proceed. Assume that the property holds only for a finite set of primes $p_1,p_2,dots, p_n$ then let
          $x:=p_1p_2cdots p_n-1$. It follows that for $i=1,dots,n$,
          $$N:=x^10+x+1equiv 1-1+1=1 pmodp_i.$$
          Now consider $q$ be a prime which divides $N$. What may we conclude?






          share|cite|improve this answer






















          • We can also take any finite set $S$ of primes such that $3in S$ and let $x=prod_pin Sp.$.............+1
            – DanielWainfleet
            Aug 30 at 15:31














          up vote
          9
          down vote



          accepted










          Yes, a proof by contradiction is a good way to proceed. Assume that the property holds only for a finite set of primes $p_1,p_2,dots, p_n$ then let
          $x:=p_1p_2cdots p_n-1$. It follows that for $i=1,dots,n$,
          $$N:=x^10+x+1equiv 1-1+1=1 pmodp_i.$$
          Now consider $q$ be a prime which divides $N$. What may we conclude?






          share|cite|improve this answer






















          • We can also take any finite set $S$ of primes such that $3in S$ and let $x=prod_pin Sp.$.............+1
            – DanielWainfleet
            Aug 30 at 15:31












          up vote
          9
          down vote



          accepted







          up vote
          9
          down vote



          accepted






          Yes, a proof by contradiction is a good way to proceed. Assume that the property holds only for a finite set of primes $p_1,p_2,dots, p_n$ then let
          $x:=p_1p_2cdots p_n-1$. It follows that for $i=1,dots,n$,
          $$N:=x^10+x+1equiv 1-1+1=1 pmodp_i.$$
          Now consider $q$ be a prime which divides $N$. What may we conclude?






          share|cite|improve this answer














          Yes, a proof by contradiction is a good way to proceed. Assume that the property holds only for a finite set of primes $p_1,p_2,dots, p_n$ then let
          $x:=p_1p_2cdots p_n-1$. It follows that for $i=1,dots,n$,
          $$N:=x^10+x+1equiv 1-1+1=1 pmodp_i.$$
          Now consider $q$ be a prime which divides $N$. What may we conclude?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 30 at 16:04

























          answered Aug 30 at 7:49









          Robert Z

          85.5k1055123




          85.5k1055123











          • We can also take any finite set $S$ of primes such that $3in S$ and let $x=prod_pin Sp.$.............+1
            – DanielWainfleet
            Aug 30 at 15:31
















          • We can also take any finite set $S$ of primes such that $3in S$ and let $x=prod_pin Sp.$.............+1
            – DanielWainfleet
            Aug 30 at 15:31















          We can also take any finite set $S$ of primes such that $3in S$ and let $x=prod_pin Sp.$.............+1
          – DanielWainfleet
          Aug 30 at 15:31




          We can also take any finite set $S$ of primes such that $3in S$ and let $x=prod_pin Sp.$.............+1
          – DanielWainfleet
          Aug 30 at 15:31

















           

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