Can You Solve Pell's Equation?

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x^2-dy^2=1
#numbertheory #PellsEquation #diophantineequations
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[@EneriGiilaan]

Hmmmm ... what's the deal with using d here?
Usually Pell's equation is written as:
x² − ny² = 1
Where following established mathematical notation conventions n indicates an integer (positive non-square integer in this specific case).
The dy² on the other hand will be easily confused with differential of y².

[@ShortsOfSyber]

I've seen d being commonly used but I agree with you. It's confusing

[@EneriGiilaan]

Agreeable not a big deal at all - I was just first confused myself :) @@ShortsOfSyber

[@pietergeerkens6324]

Guess-and-check will take a while for
x^2 - 67*y^2 = 1.
😉
Are you familiar with this old Martin Gardner classic:
If you should happen to meet two of the Jones sisters, the odds are exactly 50% that they are both blonde.
How many Jones sisters are there (assuming reasonable limits on human reproduction)?
Then, can you see how to use Pell's equation to find solutions for species with larger broods?
To my surprise, no edition of Gardner's works that I'm familiar with gives this relationship to Pell.

[@ShortsOfSyber]

Interesting

[@birandkoray]

you did not mention how to find the basic solution
how do we get (3,2) for D = 2, i want to hear it

[@ShortsOfSyber]

continued fractions
see www.ams.org/notices/200202/fea-lenstra.pdf page 2

[@Qermaq]

So what's the deal with d=4? Why does that one get two non-prime-factorable numbers, and are there other values of d which do that?

[@pietergeerkens6324]

For any d = n^2 the equation will factor as
(x - ny) (x + ny)
with the only non-negative integer solution being (x,y) = (1,0). That's because the only two perfect squares separated by 1 are 0 and 1.
Likewise, if d has any square factor so that d = m^2 * d', then the equation reduces as
1 = x^2 - d*y^2
= x^2 - (m^2'*d) * y^2
= x^2 - d' * (m*y)^2
and has the same solutions, for (x,my), as the Pell equation for d has for (x,y). Thus only d values with no square factors are of interest.

[@ShortsOfSyber]

Thank you for the response!

[@nicolascamargo8339]

Wow