A Killer Problem By Mohammed Imran

Every pair of integers $x_n,y_n$ such that $x_n + y_n\sqrt{3} \; = \; (2 + \sqrt{3})^n$ are such that $x_n^2 - 3y_n^2 = 1$ for any $n \in \mathbb{N}$ . If we set $c_n = x_n - y_n$ and $\ell_n = x_n - 3y_n$ then $3c_n^2 - \ell_n^2 \; = \; 3(x_n - y_n)^2 - (x_n - 3y_n)^2 \; = \; 2x_n^2 - 6y_n^2 \; = \; 2$ Now we note that $\begin{aligned} c_n\sqrt{3} - \ell_n & = \; (\sqrt{3}-1)(2 +\sqrt{3})^n \\ -c_n\sqrt{3} - \ell_n & = \; -(\sqrt{3}+1)(2 - \sqrt{3})^n \\ 2c_n\sqrt{3} & = \; (\sqrt{3}-1)(2 + \sqrt{3})^n + (\sqrt{3}+1)(2 - \sqrt{3})^n \\ 2(c_n -c_{n-1})\sqrt{3} & = \; (\sqrt{3}-1)(1 + \sqrt{3})(2 + \sqrt{3})^{n-1} + (\sqrt{3}+1)(1 - \sqrt{3})(2 - \sqrt{3})^{n-1} \; = \; 2\left[(2 + \sqrt{3})^{n-1} - (2 - \sqrt{3})^{n-1}\right] \end{aligned}$ and hence $c_n > c_{n-1}$ for all $n \ge 2$ . Since $c_0=c_1 = 1$ we deduce that $c_n > 1$ for all $n \ge 2$ , and hence there are infinitely many solutions in integers to the equation $3c^2 - 2 = \ell^2$ , with $c > 1$ .