# Slight Alteration to a Diophantine Result

Hello all!

In the pursuit of a minor research problem I was pointed in the direction of an interesting result in the realm of Diophantine Analysis. The content of the result follows:

$\frac{ax^{n+2l}-1}{ax^{n}-1} = y^2$ has a solution in the natural numbers for $a$, $x$, $n$, and $l$, with $x>1$ and $y$ a rational number, if and only if the following conditions are met:

• $2|l$
• $a = \frac{3^{l-1}+1}{4}$
• $x=3$
• $n=1$
• $y=\pm (3^l +2)$

I am looking to make a slight alteration to this theorem for use in my research (though the term should be interpreted lightly), by changing $y^2$ to $3y^2$. However, after conferring with one of my professors, I was told such a theorem may already exist!

Does anyone know of this (or these) results, and would be willing to suggest a reference? If no such theorem exists, does anyone have any pointers on a good approach to this problem?

-Richard Voepel

[EDIT]

I first read this result from one of my professor's papers, though I was told directly this was not the first paper to provide such a result. Here is a link.

http://www.math.sc.edu/~filaseta/papers/DiophantinePaper2006.pdf

-
It's not clear to me whether the first result you mention is yours or something you found in print. If the latter, it would help if you shared the source. In the mean time, you could see if there's anything in Mordell's Diophantine Equations, or in Guy's Unsolved Problems In Number Theory. Certainly the equation $y^2=(x^n-1)/(x-1)$ has a literature. –  Gerry Myerson Apr 15 '11 at 2:16
I apologize for the oversight! I did not develop this result in the slightest, and should have made that more clear. An edit has been made to the original post explaining this, and providing a link to the first paper I found it in. I'll make more changes in the future as I find out more information from my professor. –  Richard Voepel Apr 17 '11 at 6:18
You should work filaseta paper... –  Luis H Gallardo Apr 21 '11 at 0:27