Quadratic Number Fields and Integer Solutions

Date: 08/16/2007 at 08:01:19
From: ziya
Subject: my question is about the number theory 

Prove that the equation 34*y^2 - x^2 = 1 in Z (integer number set) has 
no answer.  I have tested different module but it has no result.

Date: 08/16/2007 at 21:19:01
From: Doctor Vogler
Subject: Re: my question is about the number theory

Hi Ziya,

Thanks for writing to Dr. Math.  You can't prove that it doesn't have
integer solutions using modular arithmetic since it DOES have rational
solutions.  For example,

  34(1/3)^2 - (5/3)^2 = 1,

and

  34(1/5)^2 - (3/5)^2 = 1.

Therefore, mod p^k for any prime p, you can invert either 3 or 5 (or
usually both) and get solutions mod p^k.  Then the Chinese Remainder
Theorem can be used to get solutions mod m for ANY positive integer m.

The only way I can think of to prove that your equation has no integer
solutions is to use the theory of quadratic number fields.  But you
can still prove that it has no integer solutions using some algebraic 
number theory.

Someone familiar with number fields would write your equation as

  x^2 - 34 y^2 = -1

and would factor it in the the ring of algebraic integers of the form

  a + b*sqrt(34)

as

  (x - y*sqrt(34))(x + y*sqrt(34)) = -1,

or

  Norm(x + y*sqrt(34)) = -1.

Such a solution would give two "units" in this ring, and there is a
theorem (Dirichlet's Unit Theorem) which says that all units in this
ring are powers of a single number (called the "Fundamental Unit")
possibly times -1.  There are also algorithms for computing the
fundamental unit of a ring like Z[sqrt(N)], and for your ring it turns
out to be

  35 + 6*sqrt(34).

When the norm of the fundamental unit is -1, its square has norm +1,
and half of the units will give a solution to

  x^2 - N y^2 = -1

and the other half will give a solution to

  x^2 - N y^2 = 1.

But 35 + 6*sqrt(34) has norm +1 (not -1), and when this happens, all
units give solutions to the equation

  x^2 - N y^2 = 1,

which means that you can never get -1 on the right.  There are plenty
of N's for which this is true, and there are some theorems stating
that certain kinds of N will always have fundamental units of a
particular norm, but there are other kinds of N where you just have to
compute the fundamental unit and calculate its norm.

Algebraic number theory (and number fields) is a fascinating subject
with a lot of very nice theorems.  If you are interested in the
subject, you could start by reading "Number Fields" by Marcus,
although there are plenty of other good books on the subject as well.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/