Consider the equation $y^2-3x^2 = -2$. It can be factorized as $(y-\sqrt3 x)(y+\sqrt3 x) = -2.$
Now, if we put $(y+\sqrt3 x) = (y_0+\sqrt3 x_0) (2+\sqrt3)^m ; m \geq0.$
It is easily verified (by combining this equation with its conjugate) that $x_0$ is always positive but that $y_0$ is negative if $m$ is large.
I am unable to understand how can one conclude the bold part.
After combining with the conjugate I got $-2 = (y^2_0-3x^2_0)(1)^m,$ which does not lead to any conclusion.
Any sort of help is appreciated.
Here is the link for the pdf file of the manuscript https://academic.oup.com/qjmath/article/20/1/129/1539395
It is assumed that $x$ and $y$ are positive integers. Then $y + \sqrt{3}\,x > 0$. Since $2 + \sqrt{3} > 1$, we have $$0 < y_0 + \sqrt{3}\,x_0 = \frac{y + \sqrt{3}\,x}{(2 + \sqrt{3})^m} < 1 \tag{1}$$ for large enough $m$. Neither of $x_0$ and $y_0$ can be zero since $y_0^2 - 3 x_0^2 = -2$. It follows that for $m$ so large that $(1)$ holds, $x_0$ and $y_0$ must have opposite signs. For if both were negative then $y_0 + \sqrt{3}\,x_0 < 0$, and if both were positive then $y_0 + \sqrt{3}\,x_0 > 1$. Thus $y_0$ and $-x_0$ have the same sign. Since $$y_0 - \sqrt{3}\,x_0 = \frac{-2}{y_0 + \sqrt{3}\,x_0} < 0\,,$$ that sign must be negative, i.e. $y_0 < 0$ and $-x_0 < 0$.