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Integer Proof Using Diophantine Equation

Date: 01/10/2005 at 10:38:58
From: Eugene
Subject: Diophantine equations??

The integer 26 is the only integer preceded by a a square and followed
by a cube.  What is the proof of this?



Date: 01/10/2005 at 11:32:06
From: Doctor Vogler
Subject: Re: Diophantine equations??

Hi Eugene,

Thanks for writing to Dr Math.  I'm surprised you recognized this as a
Diophantine equation, but you're absolutely right.  You are looking
for integer solutions to

  n + 1 = x^3
  n - 1 = y^2,

which, of course, is the same as solving

  y^2 + 1 = x^3 - 1

or

  y^2 = x^3 - 2.

This last is, in fact, an elliptic curve, and so there are methods to
find all rational solutions as well.  But you wanted all of the
integer solutions.  For this, you can use the theory of number fields.
This does, unfortunately, require some background knowledge.  I shall
give a description of the solution using some theory from number 
fields, but you might find some gaps in it if you are unfamiliar with
the subject.

First we change the equation to

  x^3 = y^2 + 2

and we wish to factor the right side.  Of course, the right side does
not factor over the rational numbers (or the real numbers), but it
does in the field of rational numbers with sqrt(-2), which I shall call

  a = sqrt(-2),

so that the field is Q(a).

Then we get

  x^3 = y^2 + 2 = (y - a)(y + a).

We want to show that the two factors on the right are relatively 
prime.  If we can do that, then the knowledge that the ring of 
integers in Q(a) is a PID and therefore a UFD will tell us that

  y - a  and  y + a

are both cubes times some unit.  But since Q(a) is an imaginary
quadratic field, the Dirichlet Unit Theorem tells us that the only
units are +1 and -1, so that y - a and y + a are both cubes.

So let's show that y - a and y + a are relatively prime.  Clearly,
their GCD must divide their difference, which is

  -2a = a^3.

Since a is a prime number in the ring of integers of Q(a), the GCD
must be a power of a.  Now, if y is even, then 4 divides y^2, so

  x^3 = y^2 + 2

must be even but not divisible by 4, which is impossible for the cube
of an integer.  Therefore we may conclude that y is odd.  But the
prime a lies over the rational prime 2, which means that a does not
divide y, and therefore a does not divide y - a or y + a.  Therefore,
their GCD must be 1.

Now we know that y - a and y + a are both cubes of algebraic integers
in Q(a).  But the algebraic integers in Q(a) are the numbers

  r + s*a

where r and s are (normal) integers.  So that means that

  y + a = (r + s*a)^3
        = r^3 + 3r^2*sa + 3r(sa)^2 + (sa)^3
        = r^3 + 3r^2*sa + 3rs^2*(-2) + s^3*(-2)a
        = (r^3 - 6rs^2) + (3sr^2 - 2s^3)a.

But if the second part,

  s(3r^2 - 2s^2) = 1,

then that means that s is +1 or -1, and therefore

  3r^2 - 2 = +1 or -1

(in fact, equals s), and therefore r is +1 or -1 as well.  So that
means that

  y + a = (-1 + a)^3

or

  y + a = (1 + a)^3.

Solving these gives y = +5 and y = -5, respectively.  So these are the
only solutions, and they both give x^3 = 27 and therefore x = 3.

You'll see where I used some facts from algebraic number theory--that
is, the theory of number fields.  To understand those parts better, I
might recommend reading a book about number fields and learning the
subject, which I found to be very interesting.  I learned from "Number
Fields" by D.A. Marcus (which you can find by searching amazon.com for
"number fields"), and I thought it was a good book, but there are also
many other books available which teach the same kinds of things.

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/ 
Associated Topics:
College Modern Algebra
College Number Theory

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