Gurth's Conjecture

gurthbruins · 2010-05-15 22:43:14

Gurth's Conjecture:

Consider the following Diophantine Equation:

Find positive integers x, y and z
such that x^2 + 2(y^2) - z = xyz.

My conjecture is as follows:
z = 2(2^n - 1)^2 + (2^n +1)^2 and z = 2(2^n + 1)^2 + (2^n - 1)^2 always lead to solutions.

I did a lot of computing to test z's up to numbers of 40 digits or more.

I submitted my conjecture in January 1990 to David Allison, who was then the number theory specialist in the Department of Mathematics at the University of Cape Town. He took a great interest in this matter, and wrote to me (16 Jan 1990):
"I confess that I am in the same boat as you - I don't see any way to prove this."

It would seem that my conjecture remains a challenge to mathematicians : prove it, or find an example that will disprove it.

I prefer this notation for keyboard use.

ZHero · 2010-05-16 00:36:16

Hi gurthbruins!

Your conjecture seems to be a beautiful piece of hard work!!

I wish that most of the Big Brains here will try untangling this mystery!

I'd request you to tell us all a little about how you happened to stumble upon this nice conjecture.
This will give one a Path/Direction to think along; rather than having no clue as to where to start from (ye ye.. i know that most of you actually know where to start from but not everybody else does )!

gurthbruins · 2010-05-16 05:47:54

ZHero, I would suggest you start by forgetting about the conjecture and first solve the Diophantine Equation. That would be step one.

If anyone wants to do that, then next he can tackle the job of looking for more solutions. As many as possible. Along the way you will have to learn about Continued Fractions if you do not already know about them. I have seen posts on this forum dealing with Diophantine Equations and Continued Fractions - have a look at those.

I do not know if there are computer programs available these days to crunch large numbers - in 1990 there were none that I could find, I had to write my own to be able to deal with numbers of up to 800 hexadecimal digits. Using machine code for my 6502 microprocessor - computers were slower in those days.

Once you have done sufficient work actually amassing solutions to a problem, you are in a better position to spot patterns in the solutions because you understand more about how it all ticks. I would hardly call this "stumbling", rather making discoveries by thoughtful inspection coupled with understanding of what is going on.

Formulating a conjecture that fits the facts is one thing, proving it quite another. Being quite unable to do that myself, I have no tips to offer. I am not a mathematician, this is just one of my hobbies.

bobbym · 2010-05-16 06:01:48

Hi;

So you had an apple!

For computation sake I manipulated your problem into:

For n = 1,2,3, I could solve getting solutions for x,y, after that it became too difficult. There are many other solutions besides the ones generated by your 2 formulas for n. I am also not sure that it is possible to use continued fractions on this. It is not exactly a pellian. How high were you able to go with n? Can you provide that data?

gurthbruins · 2010-05-16 19:48:12

I had an apple, I enjoyed it, and I threw the core away. Meaning my side of the correspondence with D. Allison.
As far as I can remember, I got as high as 8 for n. The corresponding values for x and y went into roughly 200-500 digits, I remember they covered about half a page of foolscap.

I suppose I could enquire at the University if they still have that correspondence, but I am too lazy to do that. But if you want to enquire yourself, you have my permission to do so. You have the date of Allison's last letter, you can refer to that.

I am pretty sure there is no way to do this without Continued Fractions. I was certainly using them all the time.

Of course "There are many other solutions besides the ones generated by your 2 formulas for n." If there hadn't been, it might have been a lot easier to make my conjecture!

gurthbruins · 2010-11-10 22:59:10

Hi, bobbym... after another 6 months, I stumbled on some old notes still in my possession:
Ok, so let's go beyond n=3.
Let's do n=4 next.

Then z = 2(2^n - 1)^2 + (2^n +1)^2 ......... (just taking one of the two hypothetised values)
= 739.
Now compute the Continued Fraction for SQR(z^2 - 8). (SQR means square root of).
My notes say: "Values of x and y may be obtained by computing the proper fraction n/d from the first c denominators of the Continued Fraction for SQR(z^2 - 8).
Then x = d - n
and y = 2.d "

Explanation of c : I think you must go on with the continued fraction until the cycle of denominators reaches the critical point where it starts to repeat itself. The value of c for z=739 should be 43.

That could be a lot of fun! Want to try it?

bobbym · 2010-11-11 06:44:53

Hi;

gurthbruins wrote:

I am pretty sure there is no way to do this without Continued Fractions. I was certainly using them all the time.

While continued fractions first demonstrated by Legendre, is the preferred way. The Indian mathematician Brahmagupta was solving them using his own method many centuries before that.

Taking your problem and working on it using the convergents of CF's ( but switching to Brahmagupta when convenient ).

Starting with n = 1 we get z = 11 for the first equation.

Taking the CF of √(11^2 - 8) we get this 10, {1,1,1,2,2,1,1,1,20}...

Using the list that is in set brackets ( they periodic sequence} and taking the convergents of that.

gurthbruins wrote:

Then x = d - n
and y = 2.d

You can see that the fifth one, 19 over 12 is the answer ( x = 7 , y = 38 , z = 11 ). That makes:

x = n - d
y = 2n

This is reversed from yours. Is yours a typo or did you use another way that switched the n and d?

Also your conjecture I have verified to n = 33 by direct computation!

gurthbruins · 2010-11-11 08:52:28

bobbym wrote:

Also your conjecture I have verified to n = 33 by direct computation!

I am amazed at your power of computation! In 1989 I could only get as far as n = 13. It is great news for me to hear the conjecture has now been verified to so high a value of n.

Is your solution for n=33 printable, or would it take up the whole of the internet? Viewers might be interested to see so monstrous a solution.

bobbym · 2010-11-11 09:25:16

Hi gurthbruins;

The problem is still computationally very difficult. The high n's are achieved through a trick.
I will post when I get to n = 50. Also there is a little light concerning proving for any n.

For n = 50:

gurthbruins · 2010-11-11 21:13:31

Hi bobbym.

n up to 50 already! Absolutely amazing. Beautiful work!
I've copied these 3000 digit values for x and y into a Notepad file - fortunately it is capable of word wrap. (unlike this page). So now you can restore normality to the page.

bobbym · 2010-11-11 21:36:37

Hi;

Glad to help.

Math Is Fun Forum

#1 2010-05-15 22:43:14

Gurth's Conjecture

#2 2010-05-16 00:36:16

Re: Gurth's Conjecture

#3 2010-05-16 05:47:54

Re: Gurth's Conjecture

#4 2010-05-16 06:01:48

Re: Gurth's Conjecture

#5 2010-05-16 19:48:12

Re: Gurth's Conjecture

#6 2010-11-10 22:59:10

Re: Gurth's Conjecture

#7 2010-11-11 06:44:53

Re: Gurth's Conjecture

#8 2010-11-11 08:52:28

Re: Gurth's Conjecture

#9 2010-11-11 09:25:16

Re: Gurth's Conjecture

#10 2010-11-11 21:13:31

Re: Gurth's Conjecture

#11 2010-11-11 21:36:37

Re: Gurth's Conjecture

Board footer