# Prove that $0.a_1b_1a_2b_2a_3b_3 \ldots = 0.21742615\ldots$ is irrational, where $(a_i,b_i)$ are the solutions of $x^2-3y^2=1$

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 $k$?

Hints: If $k$ is rational then its decimal expansion repeats after some point. Let $p$ be the period, and let $N=10^p$. Then there is a finite set $F$ of pairs such that $$a_i,b_i\in\{(r+sN^n)/(N-1)\mid(r,s)\in F,\,n\geq0,\,s>0\}$$ for $i$ sufficiently large. For some fixed $(r,s),(t,u)\in F$ with $s,u>0$, there are infinitely many $n$ for which $$\left(\frac{r+sN^n}{N-1}\right)^2-3\left(\frac{t+uN^n}{N-1}\right)^2=1.$$ Now expand: $$(r^2-3t^2)+2(rs-3tu)N^n+(s^2-3u^2)N^{2n}=(N-1)^2$$ and equate coefficients of powers of $N^n$: $$r^2-3t^2=(N-1)^2,\hspace{10mm}rs-3tu=s^2-3u^2=0.$$ Then $$s^2(N-1)^2=s^2r^2-3s^2t^2=9t^2u^2-9t^2u^2=0,$$ a contradiction.
• Can you explain how there is a finite set $F$ of pairs such that $$a_i,b_i \in \{(r+sN^n)/(N-1) \mid (r,s) \in F, \,n \geq 0\}?$$ Aug 17, 2017 at 0:24
• @user19405892 If $c$ is the $p$-digit repeated number, then after reaching the repeating part each $a_i$ must be of the form $c_kcc\ldots ccc_l'$ where $c_k$ is the last $k$ digits of $c$ and $c_l'$ the first $l$. This equals $10^lN^nc_k+10^l(N^n-1)/(N-1)c+c_l'$. Aug 17, 2017 at 3:18
• I get that it equals $10^{l+pn}c_k+10^l \cdot \dfrac{10^{pn}-1}{10^p-1}c+c_l'$. Aug 18, 2017 at 2:45
• @user19405892 Sorry there was a typo in my answer; $N$ should be $10^p$ instead of $10^p-1$. Does it make sense now? Aug 18, 2017 at 3:57
• Yes, can you elaborate on the part about equating the coefficients of powers of $N^n$? Aug 18, 2017 at 4:01