Solve $an^2+bn+c=k^2$ over the integers.












2












$begingroup$


I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations







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    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09
















2












$begingroup$


I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09














2












2








2





$begingroup$


I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations







share|cite|improve this question









$endgroup$




I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations




diophantine-equations






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asked Jan 4 at 20:17









SmileyCraftSmileyCraft

3,561518




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  • 1




    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09














  • 1




    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09








1




1




$begingroup$
math.stackexchange.com/questions/2424587/…
$endgroup$
– individ
Jan 5 at 5:09




$begingroup$
math.stackexchange.com/questions/2424587/…
$endgroup$
– individ
Jan 5 at 5:09










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$begingroup$

Complete the square, and you get
$$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
or $$ (5n+7)^2 - 44 = 5 k^2 $$
With $x = 5n+7$, this is a Pell-type equation
$$ x^2 - 44 = 5 k^2 $$
where you only want solutions where $x equiv 2 mod 5$.






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    $begingroup$

    Complete the square, and you get
    $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
    or $$ (5n+7)^2 - 44 = 5 k^2 $$
    With $x = 5n+7$, this is a Pell-type equation
    $$ x^2 - 44 = 5 k^2 $$
    where you only want solutions where $x equiv 2 mod 5$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Complete the square, and you get
      $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
      or $$ (5n+7)^2 - 44 = 5 k^2 $$
      With $x = 5n+7$, this is a Pell-type equation
      $$ x^2 - 44 = 5 k^2 $$
      where you only want solutions where $x equiv 2 mod 5$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Complete the square, and you get
        $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
        or $$ (5n+7)^2 - 44 = 5 k^2 $$
        With $x = 5n+7$, this is a Pell-type equation
        $$ x^2 - 44 = 5 k^2 $$
        where you only want solutions where $x equiv 2 mod 5$.






        share|cite|improve this answer









        $endgroup$



        Complete the square, and you get
        $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
        or $$ (5n+7)^2 - 44 = 5 k^2 $$
        With $x = 5n+7$, this is a Pell-type equation
        $$ x^2 - 44 = 5 k^2 $$
        where you only want solutions where $x equiv 2 mod 5$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 20:32









        Robert IsraelRobert Israel

        326k23215469




        326k23215469






























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