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Toni Mhax
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Recently I came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation $$ x^2-\left(w^2(2^{n-2}p)^2+2^n(2^{n-2}p)\right)y^2=1,\qquad n\geq 2, $$ where all variables are non zero relatively prime integers. in $ \mathbb{Z}^*$

Is it a known case in terms of the theory? As one can read, for instance, in the notes

  • Michel Waldschmidt, Pell’s equation, author pdf.

In fact the solutions are rather elementary ideas.

Thank you.

Recently I came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation $$ x^2-\left(w^2(2^{n-2}p)^2+2^n(2^{n-2}p)\right)y^2=1,\qquad n\geq 2, $$ where variables are non zero relatively prime integers.

Is it a known case in terms of the theory? As one can read, for instance, in the notes

  • Michel Waldschmidt, Pell’s equation, author pdf.

In fact the solutions are rather elementary ideas.

Thank you.

Recently I came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation $$ x^2-\left(w^2(2^{n-2}p)^2+2^n(2^{n-2}p)\right)y^2=1,\qquad n\geq 2, $$ where all variables are in $ \mathbb{Z}^*$

Is it a known case in terms of the theory? As one can read, for instance, in the notes

  • Michel Waldschmidt, Pell’s equation, author pdf.

In fact the solutions are rather elementary ideas.

Thank you.

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David Roberts
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Recently iI came up with a positive solution  $((x,y)\neq (\pm 1;0))$ to this diophantine equation

$$x^2-(w^2(2^{n-2}p)^2+2^n(2^{n-2}p))y^2=1$$

$n\ge 2$ where $$ x^2-\left(w^2(2^{n-2}p)^2+2^n(2^{n-2}p)\right)y^2=1,\qquad n\geq 2, $$ where variables are non zero relativerelatively prime integers.

Is it a known case in terms of the theory? As one can read, for instance, in this filethe notes

https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/BamakoPell2010.pdf

  • Michel Waldschmidt, Pell’s equation, author pdf.

In fact the solutions are rather elementary ideas.

Thank you.

Recently i came up with a positive solution  $((x,y)\neq (\pm 1;0))$ to this diophantine equation

$$x^2-(w^2(2^{n-2}p)^2+2^n(2^{n-2}p))y^2=1$$

$n\ge 2$ where variables are non zero relative integers.

Is it a known case in terms of the theory? As one can read in this file

https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/BamakoPell2010.pdf

In fact the solutions are rather elementary ideas.

Thank you.

Recently I came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation $$ x^2-\left(w^2(2^{n-2}p)^2+2^n(2^{n-2}p)\right)y^2=1,\qquad n\geq 2, $$ where variables are non zero relatively prime integers.

Is it a known case in terms of the theory? As one can read, for instance, in the notes

  • Michel Waldschmidt, Pell’s equation, author pdf.

In fact the solutions are rather elementary ideas.

Thank you.

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Toni Mhax
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Recently i came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation

$$x^2-(w^2(2^{n-2}p)^2+2^n(2^{n-2}p))y^2=1$$

$n\ge 2$ where variables are non zero relative integers.

Is it a known case in terms of the theory? As one can read in this file http://www.math.jussieu.fr/∼miw/articles/pdf/BamakoPell2010.pdf

https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/BamakoPell2010.pdf

In fact the solutions are rather elementary ideas.

Thank you.

Recently i came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation

$$x^2-(w^2(2^{n-2}p)^2+2^n(2^{n-2}p))y^2=1$$

$n\ge 2$ where variables are non zero relative integers.

Is it a known case in terms of the theory? As one can read in this file http://www.math.jussieu.fr/∼miw/articles/pdf/BamakoPell2010.pdf

In fact the solutions are rather elementary ideas.

Thank you.

Recently i came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation

$$x^2-(w^2(2^{n-2}p)^2+2^n(2^{n-2}p))y^2=1$$

$n\ge 2$ where variables are non zero relative integers.

Is it a known case in terms of the theory? As one can read in this file

https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/BamakoPell2010.pdf

In fact the solutions are rather elementary ideas.

Thank you.

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Toni Mhax
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Toni Mhax
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Toni Mhax
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