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Problem:

If $x,y \in \mathbb{Z}$ satisfy $x^2-2y^2=1$, show that $6$ is the largest number always dividing $xy$

My Attempt:

I've been able to show that 6 indeed always divides $xy$, but not that it's the largest number to do so.

We have, $$\begin{align*} 2y^2 &= x^2-1 \\ \implies 2y^2 &= (x-1)(x+1) \\ \implies 4&\mid 2y^2\\ \implies 2&\mid y \\ \implies 2&\mid xy \end{align*}$$

Further, $y^2 \equiv_3 1 \implies x^2 \equiv_3 0$, hence, $3\mid xy$. Hence, $6 \mid xy$.

The general solution to the given Diophantine equation is discussed here, but I don't see how that helps us here. Any help or hints are appreciated.

Aaratrick
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    The classic way of doing this would be to find some small solutions for which $6$ was the largest number - clearly $6$ persists. In lucky cases there will be a solution with a product of $6$. – Mark Bennet Jul 10 '21 at 20:51

1 Answers1

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The maximality of $6$ follows because $3^2-2\cdot 2^2=1$ is a solution to the equation.

Arthur
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