On Two Exponents of Approximation Related to a Real Number and Its Square

http://dx.doi.org/10.4153/CJM-2007-009-3
Canad. J. Math. 59(2007), 211-224

  Published:2007-02-01
 Printed: Feb 2007

For each real number $\xi$, let $\lambdahat_2(\xi)$ denote the supremum of all real numbers $\lambda$ such that, for each sufficiently large $X$, the inequalities $|x_0| \le X$, $|x_0\xi-x_1| \le X^{-\lambda}$ and $|x_0\xi^2-x_2| \le X^{-\lambda}$ admit a solution in integers $x_0$, $x_1$ and $x_2$ not all zero, and let $\omegahat_2(\xi)$ denote the supremum of all real numbers $\omega$ such that, for each sufficiently large $X$, the dual inequalities $|x_0+x_1\xi+x_2\xi^2| \le X^{-\omega}$, $|x_1| \le X$ and $|x_2| \le X$ admit a solution in integers $x_0$, $x_1$ and $x_2$ not all zero. Answering a question of Y.~Bugeaud and M.~Laurent, we show that the exponents $\lambdahat_2(\xi)$ where $\xi$ ranges through all real numbers with $[\bQ(\xi)\wcol\bQ]>2$ form a dense subset of the interval $[1/2, (\sqrt{5}-1)/2]$ while, for the same values of $\xi$, the dual exponents $\omegahat_2(\xi)$ form a dense subset of $[2, (\sqrt{5}+3)/2]$. Part of the proof rests on a result of V.~Jarn\'{\i}k showing that $\lambdahat_2(\xi) = 1-\omegahat_2(\xi)^{-1}$ for any real number $\xi$ with $[\bQ(\xi)\wcol\bQ]>2$.