Let $S$ be the set of all ordered pairs of positive integers $(x,y)$ such that $x^2-3y^2 = 1$. Let $(a_1,b_1),(a_2,b_2),\ldots$ be the elements of $S$ listed in ascending order, i.e., $a_i < a_{i+1}$ for all $i \geq 1$. Prove that $$k = 0.a_1b_1a_2b_2a_3b_3 \ldots = 0.21742615\ldots$$ is irrational.
The solutions to the equation $x^2-3y^2 = 1$ are the solutions to $x+y\sqrt{3} = (2+\sqrt{3})^m$. Therefore, $$x = \sum_{i=0}^{\left\lfloor\frac{m}{2}\right\rfloor} 2^{m-2i} \cdot 3^i \binom{m}{2i} \quad \text{and} \quad y = \sum_{i=0}^{\left\lfloor\frac{m-1}{2}\right\rfloor}2^{m-2i-1} \cdot 3^i\binom{m}{2i+1}.$$ How can we use this to find the value of $k$?