$x^2 + 3xy + y^2 = n$ Diophantine Equation












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I was wondering if someone could direct me towards information regarding the $x^2 + 3xy +y^2 = n$ diophantine equation. Additionally, is there anything about the general case of these diophantine equations in the form $x^{2k} + 3x^ky^k + y^{2k} = n$?










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  • 4




    $begingroup$
    That's $(2x+3y)^2-5y^2=4n$, so a generalised Pell equation.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 21:08










  • $begingroup$
    While I was looking for solutions to the generalized Pell equation, I encountered solutions to equations such as $x^2 - Dy^2 = pm 4$ if $D equiv 1 pmod{4}$. The solutions to this diophantine equation are given by: $frac{x+ysqrt{5}}{2} = pm left( frac{x_0 + y_0 sqrt{D}}{2} right)^k$. How can this be extrapolated to solve the case where $x^2 - Dy^2 = 4n$?
    $endgroup$
    – OmicronGamma
    Dec 24 '18 at 23:47












  • $begingroup$
    Here's a well known quadratic diophantine equation solver. alpertron.com.ar/QUAD.HTM
    $endgroup$
    – user25406
    Dec 25 '18 at 21:28
















2












$begingroup$


I was wondering if someone could direct me towards information regarding the $x^2 + 3xy +y^2 = n$ diophantine equation. Additionally, is there anything about the general case of these diophantine equations in the form $x^{2k} + 3x^ky^k + y^{2k} = n$?










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    That's $(2x+3y)^2-5y^2=4n$, so a generalised Pell equation.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 21:08










  • $begingroup$
    While I was looking for solutions to the generalized Pell equation, I encountered solutions to equations such as $x^2 - Dy^2 = pm 4$ if $D equiv 1 pmod{4}$. The solutions to this diophantine equation are given by: $frac{x+ysqrt{5}}{2} = pm left( frac{x_0 + y_0 sqrt{D}}{2} right)^k$. How can this be extrapolated to solve the case where $x^2 - Dy^2 = 4n$?
    $endgroup$
    – OmicronGamma
    Dec 24 '18 at 23:47












  • $begingroup$
    Here's a well known quadratic diophantine equation solver. alpertron.com.ar/QUAD.HTM
    $endgroup$
    – user25406
    Dec 25 '18 at 21:28














2












2








2





$begingroup$


I was wondering if someone could direct me towards information regarding the $x^2 + 3xy +y^2 = n$ diophantine equation. Additionally, is there anything about the general case of these diophantine equations in the form $x^{2k} + 3x^ky^k + y^{2k} = n$?










share|cite|improve this question











$endgroup$




I was wondering if someone could direct me towards information regarding the $x^2 + 3xy +y^2 = n$ diophantine equation. Additionally, is there anything about the general case of these diophantine equations in the form $x^{2k} + 3x^ky^k + y^{2k} = n$?







number-theory diophantine-equations quadratic-forms pell-type-equations






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edited Dec 25 '18 at 0:13







OmicronGamma

















asked Dec 24 '18 at 21:01









OmicronGammaOmicronGamma

607




607








  • 4




    $begingroup$
    That's $(2x+3y)^2-5y^2=4n$, so a generalised Pell equation.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 21:08










  • $begingroup$
    While I was looking for solutions to the generalized Pell equation, I encountered solutions to equations such as $x^2 - Dy^2 = pm 4$ if $D equiv 1 pmod{4}$. The solutions to this diophantine equation are given by: $frac{x+ysqrt{5}}{2} = pm left( frac{x_0 + y_0 sqrt{D}}{2} right)^k$. How can this be extrapolated to solve the case where $x^2 - Dy^2 = 4n$?
    $endgroup$
    – OmicronGamma
    Dec 24 '18 at 23:47












  • $begingroup$
    Here's a well known quadratic diophantine equation solver. alpertron.com.ar/QUAD.HTM
    $endgroup$
    – user25406
    Dec 25 '18 at 21:28














  • 4




    $begingroup$
    That's $(2x+3y)^2-5y^2=4n$, so a generalised Pell equation.
    $endgroup$
    – Lord Shark the Unknown
    Dec 24 '18 at 21:08










  • $begingroup$
    While I was looking for solutions to the generalized Pell equation, I encountered solutions to equations such as $x^2 - Dy^2 = pm 4$ if $D equiv 1 pmod{4}$. The solutions to this diophantine equation are given by: $frac{x+ysqrt{5}}{2} = pm left( frac{x_0 + y_0 sqrt{D}}{2} right)^k$. How can this be extrapolated to solve the case where $x^2 - Dy^2 = 4n$?
    $endgroup$
    – OmicronGamma
    Dec 24 '18 at 23:47












  • $begingroup$
    Here's a well known quadratic diophantine equation solver. alpertron.com.ar/QUAD.HTM
    $endgroup$
    – user25406
    Dec 25 '18 at 21:28








4




4




$begingroup$
That's $(2x+3y)^2-5y^2=4n$, so a generalised Pell equation.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 21:08




$begingroup$
That's $(2x+3y)^2-5y^2=4n$, so a generalised Pell equation.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 21:08












$begingroup$
While I was looking for solutions to the generalized Pell equation, I encountered solutions to equations such as $x^2 - Dy^2 = pm 4$ if $D equiv 1 pmod{4}$. The solutions to this diophantine equation are given by: $frac{x+ysqrt{5}}{2} = pm left( frac{x_0 + y_0 sqrt{D}}{2} right)^k$. How can this be extrapolated to solve the case where $x^2 - Dy^2 = 4n$?
$endgroup$
– OmicronGamma
Dec 24 '18 at 23:47






$begingroup$
While I was looking for solutions to the generalized Pell equation, I encountered solutions to equations such as $x^2 - Dy^2 = pm 4$ if $D equiv 1 pmod{4}$. The solutions to this diophantine equation are given by: $frac{x+ysqrt{5}}{2} = pm left( frac{x_0 + y_0 sqrt{D}}{2} right)^k$. How can this be extrapolated to solve the case where $x^2 - Dy^2 = 4n$?
$endgroup$
– OmicronGamma
Dec 24 '18 at 23:47














$begingroup$
Here's a well known quadratic diophantine equation solver. alpertron.com.ar/QUAD.HTM
$endgroup$
– user25406
Dec 25 '18 at 21:28




$begingroup$
Here's a well known quadratic diophantine equation solver. alpertron.com.ar/QUAD.HTM
$endgroup$
– user25406
Dec 25 '18 at 21:28










1 Answer
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$begingroup$

I recommend Conways's topograph method. Here is the diagram for $u^2 + uv - v^2,$ which is "equivalent" to your form, which is reduced in Zagier's style.



enter image description here
enter image description hereenter image description hereenter image description here






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    1 Answer
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    1 Answer
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    active

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    2












    $begingroup$

    I recommend Conways's topograph method. Here is the diagram for $u^2 + uv - v^2,$ which is "equivalent" to your form, which is reduced in Zagier's style.



    enter image description here
    enter image description hereenter image description hereenter image description here






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I recommend Conways's topograph method. Here is the diagram for $u^2 + uv - v^2,$ which is "equivalent" to your form, which is reduced in Zagier's style.



      enter image description here
      enter image description hereenter image description hereenter image description here






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I recommend Conways's topograph method. Here is the diagram for $u^2 + uv - v^2,$ which is "equivalent" to your form, which is reduced in Zagier's style.



        enter image description here
        enter image description hereenter image description hereenter image description here






        share|cite|improve this answer









        $endgroup$



        I recommend Conways's topograph method. Here is the diagram for $u^2 + uv - v^2,$ which is "equivalent" to your form, which is reduced in Zagier's style.



        enter image description here
        enter image description hereenter image description hereenter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 21:50









        Will JagyWill Jagy

        105k5103202




        105k5103202






























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