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?

Note by John Ashley Capellan
6 years ago

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

Calvin Lin Staff - 5 years, 10 months ago

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

1) Observe that (5,2) (5,2) is the first non-trivial solution to x26y2=1 x^2 - 6y^2 = 1 .

2) Observe that

(a26b2)(c26d2)=a2c2+36b2d26b2c26a2d2=(ac+6bd)26(bc+ad)2. ( 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)(c,d)=(ac+6bd,bc+ad) (a,b) \otimes (c, d) = ( ac + 6 bd , bc + ad) .

3) Observe that (3,1) (3, 1) is a solution to x26y2=3 x^2 - 6y^2 = 3 .

4) Hence, solutions exist in the form of (3,1)(5,2)n (3,1) \otimes ( 5,2)^ n , where nn is a non-negative integer.
For example, with n=1 n=1 , we get (3,1)(5,2)=(15+12,5+6)=(27,11) (3,1) \otimes (5,2) = ( 15 + 12 , 5 + 6) = (27, 11). We can check that 2726×112=729726=3 27^2 - 6 \times 11^2 = 729 - 726 = 3 .

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

Calvin Lin Staff - 6 years ago

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You also left out an a2a^2 in Point 2. Cheers, G.

A Former Brilliant Member - 1 year, 12 months ago

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Fixed. Thanks!

Calvin Lin Staff - 1 year, 12 months ago

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

mathh mathh - 5 years, 7 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.

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

Usama Khidir - 6 years ago

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

A Former Brilliant Member - 5 years, 12 months ago

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