Is it possible to find $x$ in a general form or forms for the equation $2n^2 + 2n + 1$ mod $x = 0$?












0












$begingroup$


Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?



$2n^2 + 2n + 1$ mod$ x = 0$



ie



$x = 1,5,13,17$... work and solve the condition.






elementary-number-theory modular-arithmetic







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?



    $2n^2 + 2n + 1$ mod$ x = 0$



    ie



    $x = 1,5,13,17$... work and solve the condition.






    elementary-number-theory modular-arithmetic







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?



      $2n^2 + 2n + 1$ mod$ x = 0$



      ie



      $x = 1,5,13,17$... work and solve the condition.






      elementary-number-theory modular-arithmetic







      share|cite|improve this question











      $endgroup$




      Hello I was wondering is possible to find x in a general form or forms so that there are integer answers for something like the equation below? If so could you point me in the right direction?



      $2n^2 + 2n + 1$ mod$ x = 0$



      ie



      $x = 1,5,13,17$... work and solve the condition.






      elementary-number-theory modular-arithmetic




      elementary-number-theory modular-arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 22 '18 at 5:58









      AandJ4Ever

      114




      114










      asked Dec 22 '18 at 5:30









      Ryan ToppsRyan Topps

      124




      124






















          2 Answers
          2






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          3












          $begingroup$

          I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.



          First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:



          $$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$



          This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.



          By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.



          It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.



          In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, you explained that well
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 7:36










          • $begingroup$
            @RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
            $endgroup$
            – Slade
            Dec 22 '18 at 7:44



















          0












          $begingroup$

          Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            17 is not part of f(n) though
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 5:57






          • 1




            $begingroup$
            @RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
            $endgroup$
            – AandJ4Ever
            Dec 22 '18 at 6:01













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.



          First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:



          $$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$



          This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.



          By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.



          It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.



          In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, you explained that well
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 7:36










          • $begingroup$
            @RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
            $endgroup$
            – Slade
            Dec 22 '18 at 7:44
















          3












          $begingroup$

          I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.



          First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:



          $$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$



          This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.



          By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.



          It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.



          In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you, you explained that well
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 7:36










          • $begingroup$
            @RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
            $endgroup$
            – Slade
            Dec 22 '18 at 7:44














          3












          3








          3





          $begingroup$

          I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.



          First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:



          $$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$



          This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.



          By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.



          It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.



          In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.






          share|cite|improve this answer











          $endgroup$



          I think you're asking for what $x$ the equation $2n^2 + 2n + 1$ has a root modulo $x$.



          First, since $2n^2 + 2n + 1$ is odd, we can assume that $x$ is odd. Then we can write:



          $$2n^2 + 2n + 1 = frac{1}{2}((2n+1)^2 + 1)$$



          This allows us to see that the equation has a solution if and only if there is a square root of $-1$ modulo $x$.



          By the Chinese Remainder Theorem, this holds exactly when there is a square root of $-1$ modulo $p^k$ for every prime power $p^k$ dividing $x$.



          It's a standard result that this happens when $pequiv 1pmod{4}$. This can be seen by looking at the cyclic group of units $pmod{p^k}$ and seeing when it has order divisible by $4$.



          In conclusion, your equation has a solution exactly when all of the prime factors of $x$ are $1pmod 4$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 22 '18 at 7:43

























          answered Dec 22 '18 at 7:07









          SladeSlade

          25.1k12665




          25.1k12665












          • $begingroup$
            Thank you, you explained that well
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 7:36










          • $begingroup$
            @RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
            $endgroup$
            – Slade
            Dec 22 '18 at 7:44


















          • $begingroup$
            Thank you, you explained that well
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 7:36










          • $begingroup$
            @RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
            $endgroup$
            – Slade
            Dec 22 '18 at 7:44
















          $begingroup$
          Thank you, you explained that well
          $endgroup$
          – Ryan Topps
          Dec 22 '18 at 7:36




          $begingroup$
          Thank you, you explained that well
          $endgroup$
          – Ryan Topps
          Dec 22 '18 at 7:36












          $begingroup$
          @RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
          $endgroup$
          – Slade
          Dec 22 '18 at 7:44




          $begingroup$
          @RyanTopps I made a mistake; see the updated answer. The prime factors can never be $3pmod{4}$, I was misremembering something else.
          $endgroup$
          – Slade
          Dec 22 '18 at 7:44











          0












          $begingroup$

          Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            17 is not part of f(n) though
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 5:57






          • 1




            $begingroup$
            @RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
            $endgroup$
            – AandJ4Ever
            Dec 22 '18 at 6:01


















          0












          $begingroup$

          Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            17 is not part of f(n) though
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 5:57






          • 1




            $begingroup$
            @RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
            $endgroup$
            – AandJ4Ever
            Dec 22 '18 at 6:01
















          0












          0








          0





          $begingroup$

          Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.






          share|cite|improve this answer









          $endgroup$



          Yes: $f(n)$ mod $x=0$ if and only if $xmid f(n)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 5:54









          AandJ4EverAandJ4Ever

          114




          114












          • $begingroup$
            17 is not part of f(n) though
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 5:57






          • 1




            $begingroup$
            @RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
            $endgroup$
            – AandJ4Ever
            Dec 22 '18 at 6:01




















          • $begingroup$
            17 is not part of f(n) though
            $endgroup$
            – Ryan Topps
            Dec 22 '18 at 5:57






          • 1




            $begingroup$
            @RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
            $endgroup$
            – AandJ4Ever
            Dec 22 '18 at 6:01


















          $begingroup$
          17 is not part of f(n) though
          $endgroup$
          – Ryan Topps
          Dec 22 '18 at 5:57




          $begingroup$
          17 is not part of f(n) though
          $endgroup$
          – Ryan Topps
          Dec 22 '18 at 5:57




          1




          1




          $begingroup$
          @RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
          $endgroup$
          – AandJ4Ever
          Dec 22 '18 at 6:01






          $begingroup$
          @RyanTopps We have $amid b$ iff there is some integer $k$ s.t. $b=ak$ iff $b=0$ mod $a$.
          $endgroup$
          – AandJ4Ever
          Dec 22 '18 at 6:01




















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