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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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

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

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) – Calvin Lin Staff · 3 years, 1 month ago

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– Mathh Mathh · 2 years, 8 months ago

It's \((ac+6bd,bc+ad)\). Fix it to avoid confusion.Log in to reply

@John Ashley Capellan Can you add what you learnt about Pell's Equation to the Wiki? Thanks! – Calvin Lin Staff · 2 years, 11 months ago

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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. – Subrata Saha · 3 years, 1 month ago

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– Usama Khidir · 3 years, 1 month ago

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

– Subrata Saha · 3 years, 1 month ago

You can know about my solution by searching pell fermat equation in wikipediaLog in to reply