- I ended up with the following problem while trying to solve one of

the problems in graph theory, i felt this problem had something to do

with prime numbers and thats why i thought of primenumbers

yahoogroup :-)

Well the problem goes like this...

a(a+1) = b(b+1)(b+2)

are there any integer (positive) solutions of the above mentioned

equation....

I did try brute forcing and ended up with the answer a=14 and b=5. I

feel this is the only possibility apart from the a=1, b=2 which is

trivial.

Is 14, 5 is the only non-trivial solution or do we have many such

solutions...?

I feel this has something to do with the prime representation of

triangular numbers...

Would be happy if someone can through light on this ... :-)

Bye,

Sudarshan

Department of mathematics

PESIT college of engineering

Bangalore - 560085

Ph: +91-9886174198 - In a message dated 02/10/2004 16:53:40 GMT Daylight Time,

sudarshan@...-edu.com writes:

> a(a+1) = b(b+1)(b+2)

An interesting problem, even though rather OT.

>

> are there any integer (positive) solutions of the above mentioned

> equation....

>

> I did try brute forcing and ended up with the answer a=14 and b=5. I

> feel this is the only possibility apart from the a=1, b=2 which is

> trivial.

>

> Is 14, 5 is the only non-trivial solution or do we have many such

> solutions...?

>

> I feel this has something to do with the prime representation of

> triangular numbers...

>

> Would be happy if someone can through light on this ... :-)

>

Firstly, there are many more "trivial" solutions than the one you gave [with

a and b incorrectly interchanged].

Making the change of variables

x = 4*b+4, y = 8*a+4,

your equation becomes an elliptic equation in canonical form:-

y^2 = x^3 - 16*x + 16

which has discriminant delta = 16^3*37 > 0.

See e.g. H.Davenport "The Higher Arithmetic" (Cambridge U.P. 1999) for a good

introduction to the theory of elliptic equations and curves (which gets very

deep very quickly and eventually ends up with Andrew Wiles's proof of Fermat's

Last Theorem...)

The point P_1 (x=0, y=-4) is a rational point on this curve.

It is not a torsion point, so it generates an infinite sequence of rational

points P_n = n*P_1, where "n*" is the operation "+" of "addition" of a point of

the curve to itself n times.

If there is no other independent rational point, so that the rank of the

curve is 1, then all rational points on the curve are of this form.

The first few points P_n are as follows:-

n x y a b

1 0 -4 -1 -1

2 4 4 0 0

3 -4 4 0 -2

4 8 20 2 1

5 1 1 -3/8 -3/4

6 24 116 14 5

7 -20/9 -172/27 -35/27 -14/9

8 84/25 52/125 -56/125 -4/25

The point "-P_n" has the sign of y reversed, and is also a rational point on

the curve.

The first few of these are:-

n x y a b

1 0 4 0 -1

2 4 -4 -1 0

3 -4 -4 -1 -2

4 8 -20 -3 1

5 1 -1 -5/8 -3/4

6 24 -116 -15 5

This enumeration includes 10 integer solutions of the original equation.

If the rank of the elliptic equation is 1, then it should be possible to

/prove/ that there are no others.

Perhaps someone would like to use Pari to settle this question?

-Mike Oakes

[Non-text portions of this message have been removed] - I wrote
>Perhaps someone would like to use Pari to settle this question?

I asked David Broadhurst for help on this and, guru that he is, he has come up with the goods. To quote his email:-

>

So, the rank is indeed 1, and all rational solutions are of the form p_n = n*P_1.

>MordellWeilRank(EllipticCurve([0,3,1,2,0]));

>

>Magma V2.11-6

>Sun Oct 3 2004 11:26:59 on modular [Seed = 2793930133]

> -------------------------------------

>

> 1

>

> Total time: 0.190 seconds, Total memory usage: 3.50MB

Now "all" that remains is to show that the denominator of the x and y coordinates of P_n are never 1, for n > 6.

Ideas, anyone?

[This part is going to be /hard/: as David pointed out, the Catalan problem (a^2=b^3+1) is of the same type, and has so far resisted all attempts at a solution.]

-Mike Oakes - Not to say that this makes it "easy," but I thought the Catalan

conjecture was resolved about a year and a half ago. A Google

search found http://www.ams.org/bull/2004-41-01/S0273-0979-03-00993-

5/S0273-0979-03-00993-5.pdf as an AMS Bulletin summarizing the

history.

On the other hand I tested out to .... oh shoot, I cut and pasted

something over the cut and pasted number that I had searched up to

but it is started with a 2 and many digits, like 2 million or 2

billion. Darn darn darn. Oh well, not any more solutions for a

long long time.

Adam

--- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:

> I wrote

> >Perhaps someone would like to use Pari to settle this question?

>

> I asked David Broadhurst for help on this and, guru that he is, he

has come up with the goods. To quote his email:-

> >

> >MordellWeilRank(EllipticCurve([0,3,1,2,0]));

> >

> >Magma V2.11-6

> >Sun Oct 3 2004 11:26:59 on modular [Seed = 2793930133]

> > -------------------------------------

> >

> > 1

> >

> > Total time: 0.190 seconds, Total memory usage: 3.50MB

>

> So, the rank is indeed 1, and all rational solutions are of the

form p_n = n*P_1.

> Now "all" that remains is to show that the denominator of the x

and y coordinates of P_n are never 1, for n > 6.

>

> Ideas, anyone?

>

> [This part is going to be /hard/: as David pointed out, the

Catalan problem (a^2=b^3+1) is of the same type, and has so far

resisted all attempts at a solution.]

>

> -Mike Oakes