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Let $d > 0$ be square-free. Let $\epsilon = x_0 + y_0 \sqrt{d}$ be the minimal solution to the Pell's equation $x ^ 2 - d y ^ 2 = 1$. Let $x + y \sqrt{d} = \epsilon ^ l, l \geq 1$ be a solution.

Question: If all the prime divisors of $x$ divides $x_0$, does it follow that $l = 1$, i.e. $x + y \sqrt{d} = x_0 + y_0 \sqrt{d}$ (or in other words, $x + y \sqrt{d}$ is the minimal solution) ?

minimax
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  • what is the source of the question? i.e., why do you think this is worth investigating? – Will Jagy Mar 26 '14 at 20:37
  • I was trying to solve the Diophantine equation $a(x^n - x^m) = (a x^m - 4) y^2$ with the condition that all variables are positive integers and $ax$ is odd. One of the case can be easily deduced if the above is true. This is a problem in Chinese IMO team training. I am not sure whether it has any deeper mathematical content though. – minimax Mar 27 '14 at 03:19
  • This may represent a generalization of my question http://math.stackexchange.com/questions/827948/are-pell-solutions-unique, though you're looking at $x$, and I was considering the factors of $y$. In any case, I think there are many potential applications of this result, if it were proven. – Kieren MacMillan Oct 04 '14 at 16:17

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