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- Feb 7, 2012

- 2,681

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- Feb 7, 2012

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- Feb 7, 2012

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Hint:

- Oct 18, 2017

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As a matter of fact, I was interested in that very question. The question is about finding points with integer coordinates on a hyperbola, and this is a classical problem on representation by quadratic forms. I wrote something about it here. Sorry, it's in French, but ‶the equations speak for themselves″

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- Feb 7, 2012

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The convergents $\frac vu$ are alternately slightly larger and slightly smaller than $\sqrt6$, corresponding to solutions of $v^2 = 6u^2+1$ and $v^2 = 6u^2-1$. So from alternate rows of that table we get

$$\begin{array}{r|r|r|r}v&u&x=u(3u+v)&y=u(2u+v) \\ \hline 5&2&22&18 \\ 49&20&2180&1780 \\ 485&198&213642&174438 \\ 4801&1960&20934760&17093160\end{array}$$

After that, the numbers increase rapidly, giving an infinite sequence of positive integer solutions of $2x^2+x = 3y^2+y$. (I'm pleased to see that my numbers tally exactly with castor28 's!)