Solving Pell Equation of Norms other than -1 and 1

People, I crossed a Pell-type equation: x^2 - 6y^2 = 3 which it has norm 3. Are there ways to solve this equation without using concepts from Abstract Algebra such as factoring in a number field or what?

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As such, we define pair-multiplication as $(a,b) \otimes (c, d) = ( ac + 6 bd , bc + ad)$.

3) Observe that $(3, 1)$ is a solution to $x^2 - 6y^2 = 3$.

4) Hence, solutions exist in the form of $(3,1) \otimes ( 5,2)^ n$, where $n$ is a non-negative integer.
For example, with $n=1$, we get $(3,1) \otimes (5,2) = ( 15 + 12 , 5 + 6) = (27, 11)$. We can check that $27^2 - 6 \times 11^2 = 729 - 726 = 3$.

Followup question: Are there other solutions? (ignore negative values)

First find the smallest positive solution of x,y. Express it as x+6^0.5y
Then find solution of the equation x^2-6y^2=1. Express it as x+6^0.5y Multiply any of these two you get another solution. Hence obtain all solutions.

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## Comments

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TopNewestYou can read up on Pell's Equation, to understand how to generate solutions from a base case.

1) Observe that $(5,2)$ is the first non-trivial solution to $x^2 - 6y^2 = 1$.

2) Observe that

$( a^2 - 6b^2 ) ( c^2 - 6 d^2 ) = a^2c^2 + 36 b^2d^2 - 6 b^2 c^2 - 6a^2d^2= ( ac + 6bd) ^2 - 6 ( bc+ad) ^2.$

As such, we define pair-multiplication as $(a,b) \otimes (c, d) = ( ac + 6 bd , bc + ad)$.

3) Observe that $(3, 1)$ is a solution to $x^2 - 6y^2 = 3$.

4) Hence, solutions exist in the form of $(3,1) \otimes ( 5,2)^ n$, where $n$ is a non-negative integer.

For example, with $n=1$, we get $(3,1) \otimes (5,2) = ( 15 + 12 , 5 + 6) = (27, 11)$. We can check that $27^2 - 6 \times 11^2 = 729 - 726 = 3$.

Followup question: Are there other solutions? (ignore negative values)

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It's $(ac+6bd,bc+ad)$. Fix it to avoid confusion.

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You also left out an $a^2$ in Point 2. Cheers, G.

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Fixed. Thanks!

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@John Ashley Capellan Can you add what you learnt about Pell's Equation to the Wiki? Thanks!

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First find the smallest positive solution of x,y. Express it as x+6^0.5y Then find solution of the equation x^2-6y^2=1. Express it as x+6^0.5y Multiply any of these two you get another solution. Hence obtain all solutions.

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Could you please explain in a bit more detail? Sorry for the trouble.

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You can know about my solution by searching pell fermat equation in wikipedia

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