Why Can't d Be a Perfect Square in Pell Equations?

  • Thread starter Thread starter cragar
  • Start date Start date
cragar
Messages
2,546
Reaction score
3

Homework Statement


A pell equation is an equation x^2-dy^2=1 where d is a positive integer that is not a perfect square. Can you figure out why we do not want d to be a perfect square?

The Attempt at a Solution


if d was a perfect square then we would have
x^2-d^2y^2=1 z=dy then x^2-z^2=1=(x+z)(x-z)=1
x>z for this to work so if x>z then x-z is at least 1 and then x+z would be bigger than 1
so (x+z)(x-z)>1 so this won't work so d can't be a perfect square. Does this work?
 
Physics news on Phys.org
hi cragar! :wink:

yes, that's ok :smile:

(though i'd be inclined to say that the only divisors of 1 are 1 and 1,
or -1 and -1, so x+z = x-z, so z = 0 and x = 1)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top