A note on the simultaneous Pell equations \(x^2-ay^2=1\) and \(z^2-by^2=1\) (Q2901983)

Let us consider the simultaneous Pell equations NEWLINE\[NEWLINEx^2 -ay^2=1,\quad z^2 -by^2=1.NEWLINE\]NEWLINE In 2006, \textit{M. A. Bennett} et al., proved that this system has at most \(2\) solutions, see [Acta Arith. 122, No. 4, 407--417 (2006; Zbl 1165.11034)]. This is the best result one can have. In [J. Reine Angew. Math. 498, 173--199 (1998; Zbl 1044.11011)], \textit{M. A. Bennett} showed that if this system has solutions then it is equivalent to the following system of Pell equations NEWLINE\[NEWLINEX^2 -(m^2-1)Y^2=1,\quad Z^2 -(n^2-1)Y^2=1,NEWLINE\]NEWLINE where \(m, n\) are distinct positive integers such that \(\min(m, n) >1\).NEWLINENEWLINENEWLINEIn the paper under review, the author proves that if \(\delta\) is a positive number such that \(\frac{1}{2} < \delta <1\), \(\gcd(m, n)>\max(m^{\delta}, n^{\delta})\), and NEWLINE\[NEWLINE\max(m, n) > \left(\frac{8\times 10^{16}}{\theta^3} \log^3\left(\frac{10^{16}}{\theta^3}\right)\right)^{1/\theta}NEWLINE\]NEWLINE with \(\theta=\min(1-\delta, 2\delta -1)\), then the second system has only one solution \((X, Y, Z)=(m, 1, n)\). The method used for the proof is based on some properties of Lucas sequences.