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# 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
3 years, 1 month 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 $$(5,2)$$ 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) Staff · 3 years, 1 month ago

It's $$(ac+6bd,bc+ad)$$. Fix it to avoid confusion. · 2 years, 8 months ago

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

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. · 3 years, 1 month ago

Could you please explain in a bit more detail? Sorry for the trouble. · 3 years, 1 month ago