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Solving a Diophantine Equation by Use of Elliptic Curves

Date: 03/23/2008 at 17:36:33
From: Wayne
Subject: source for solution to particular Diophantine equation

Please help me to find a source for solving the equation (or similar) 
4u^3 - v^2 = 3.  It has solutions: (1,-1), (1,1), (7,-37), and (7,37).
My guess is that these are all the solutions, but I can't find any
exposition on solving these types of equations.

I've tried factoring in Z[omega], the ring of algebraic number of the 
quadratic number field obtained by adjoining the square root of 
negative 3 to the rationals.  I have not been able to make any 


Date: 04/03/2008 at 20:15:29
From: Doctor Vogler
Subject: Re: source for solution to particular diophantine equation

Hi Wayne,

Thanks for writing to Dr. Math.  I'm impressed at your trying to work
in Z[omega].  This is actually a useful method, and I do some of this
in this answer:

  Solving a Nonlinear Diophantine Equation

But often the end result is a thing known as a Thue Equation, which
takes some additional work to solve.  Either that, or you need to use
linear forms in logarithms, which are also pretty complicated.

There are, in fact, many expositions on solving these types of
equations.  The trick is to search for the phrase "elliptic curve."

Your equation v^2 = 4u^3 - 3 is equivalent to

  y^2 = x^3 - 48

where y = 4v and x = 4u.  This last equation is an elliptic curve in
standard (Weierstrass) form.

Many elliptic curves (in fact, about half of them) have only finitely
many rational solutions.  And finding the rational solutions is
generally much easier than finding the integer solutions.  So if there
are only finitely many rational solutions, then you just check which
ones are integer solutions.  I describe how to find the rational
solutions in these answers:

  Solving System of Equations Using Elliptic Curves

  Using Elliptical Curves to Solve an Arithmetic Sequence

Unfortunately for you, there are infinitely many rational solutions,
since your curve has trivial torsion and rank 1.  That is, all integer
multiples of the point (x, y) = (4, 4) are rational points on your
curve (and all rational points are integer multiples of that point). 
These multiples include

  -3(4, 4) = (73/9, -595/27)
  -2(4, 4) = (28, 148)
  -1(4, 4) = (4, -4)
   0(4, 4) = 0
   1(4, 4) = (4, 4)
   2(4, 4) = (28, -148)
   3(4, 4) = (73/9, 595/27)
   4(4, 4) = (9772/1369, -899996/50653)
   5(4, 4) = (1184884/32041, 1289162164/5735339)
   6(4, 4) = (48833569/12744900, -130710951503/45499293000)

and so on.  (I computed the rank in mwrank, and the above points in
Pari, as described in the above answers.)  The (u, v) values you can
get by dividing x and y by 4.

But every elliptic curve has only finitely many integer solutions
(this is known as Siegel's Theorem), so when there are infinitely many
rational points, it can be hard to be sure that you have found all of
the integer ones.

There are three general methods for doing this, all of them described
in detail in the book "The Algorithmic Resolution of Diophantine
Equations" by Nigel P Smart.  But all three methods are quite
complicated and take a significant amount of computation, enough to
make anyone do it on a computer.  But you can find all integer
solutions and *prove* that you have them all.

Of course, if you don't need a complete proof, then you use the fact
that the number of digits in the numerators *and* denominators of the
x and y coordinates of the point n(4, 4) is roughly a multiple of n^2
(and this becomes less rough as n gets large), which means that you're
not likely to get a denominator of 1 (i.e. an integer solution) when n
is remotely large.  So you check the first small integers like the
ones I already did, and when you see the denominator get big and keep
growing, you can be pretty confident that you aren't going to find any
more integers.

offer further suggestions.

- Doctor Vogler, The Math Forum
Associated Topics:
College Number Theory

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