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.

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!

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!

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:

Hi;

Glad to help.