Associated Topics || Dr. Math Home || Search Dr. Math

### Solving Diophantine Equations

```Date: 04/18/2008 at 14:26:34
From: Garrett
Subject: Solutions to 7x^2 + 1 = y^3 over integers (possibly reals).

I have recently delved into the world of Diophantine equations,
elliptic curves, and number fields. Looking at the equation

7x^2 + 1 = y^3

I have computationally found integer solutions (x>=0):

(x,y) = (0,1), (1,2), (3,4), and (39,22)

but hit round-off issues around x = 1.05*10^6. Still, being able to
search beyond 10^6 makes me question if these are the only integer
solutions.  Could you please answer the questions as to the number of
integer solutions to this equation and either provide an upper bound
for finite solutions (or find all solutions) or how to generate
infinite solutions?  Could you also address this question in x,y over
reals?

methods used to solve these types of equations.  I am hoping another
solution of this type of question will help me to better understand
how to go about solving these types of questions.  Thanks for the
help!

```

```
Date: 04/18/2008 at 21:23:57
From: Doctor Vogler
Subject: Re: Solutions to 7x^2 + 1 = y^3 over integers (possibly reals).

Hi Garrett,

Thanks for writing to Dr. Math.  You are correct that you have an
elliptic curve.  Finding the real points is easy.  For every real
number x, you can set y = (7x^2 + 1)^(1/3).  Alternately, y has to be
at least 1, since 7x^2 >= 0, but whenever y >= 1, there are two
solutions for x, namely sqrt((y^3 - 1)/7) and -sqrt((y^3 - 1)/7).  You
can plot this curve on a graphing calculator or computer graphing
program to see what it looks like.

Finding all of the integer points is *much* harder.  There are often
easy methods to prove that there are no integer points on such an
equation, but these methods don't work when there are some.  Methods
for finding all integer points on an elliptic curve use much of the
structure of an elliptic curve.

should put it in standard (Weierstrass) form.  If you set

v = 49x
u = 7y

and you multiply your equation by 7^3 = 343, then you get

v^2 = u^3 - 343

which is an elliptic curve in standard 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

Solving System of Equations Using Elliptic Curves
http://mathforum.org/library/drmath/view/70511.html

Using Elliptical Curves to Solve an Arithmetic Sequence
http://mathforum.org/library/drmath/view/69827.html

Unfortunately for you, there are infinitely many rational solutions,
since your curve has rank 1, and torsion of order 2.  That is, all
integer multiples of the point (u, v) = (14, 49), along with all such
points plus the torsion point (7, 0) are rational points on your
curve (and this gives you all rational points).  These multiples include

.
.
.
-5(14, 49)          = (3397814/477481, -1374504341/329939371)
-5(14, 49) + (7, 0) = (10082548/7921, 32015140437/704969)
-4(14, 49)          = (6512/169, -523921/2197)
-4(14, 49) + (7, 0) = (62146/5329, 13715247/389017)
-3(14, 49)          = (763/9, 21070/27)
-3(14, 49) + (7, 0) = (889/100, -18963/1000)
-2(14, 49)          = (8, 13)
-2(14, 49) + (7, 0) = (154, -1911)
-1(14, 49)          = (14, -49)
-1(14, 49) + (7, 0) = (28, 147)
0(14, 49)          = 0
0(14, 49) + (7, 0) = (7, 0)
1(14, 49)          = (14, 49)
1(14, 49) + (7, 0) = (28, -147)
2(14, 49)          = (8, -13)
2(14, 49) + (7, 0) = (154, 1911)
3(14, 49)          = (763/9, -21070/27)
3(14, 49) + (7, 0) = (889/100, 18963/1000)
4(14, 49)          = (6512/169, 523921/2197)
4(14, 49) + (7, 0) = (62146/5329, -13715247/389017)
5(14, 49)          = (3397814/477481, 1374504341/329939371)
5(14, 49) + (7, 0) = (10082548/7921, -32015140437/704969)
.
.
.

and so on.  (I computed the rank in mwrank, and the above points in
Pari, as described in the above answers I linked.)  The (x, y) values
you can get by dividing v by 49 and u by 7.

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.

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 Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search