Integer Sequences

Problem : Let $b$ be an integer greater than $4$ and define the number $ x_n = \underbrace{11\dots1}_{n-1} \  \underbrace{22\dots2}_{n} \ 5 $ in base $b$. Prove that $x_n $ is a perfect square for all sufficiently large $n$ if and only if $ b = 10 $.

(Source : IMO 2003 Shortlist, Laurentiu Panaitopol)

Solution :

 $\begin{align*}  x_n  & = \underbrace{11\dots1}_{n-1} \ \underbrace{22\dots2}_{n} \ 5 \\  & = \frac{b^{2n}+b^{n+1}+3b -5 }{b-1} \end{align*} $

Now the problem condition implies that $ \exists \ (y_n) \in \mathbb{Z} $ such that $ y_n^2 = x_n \ \forall \ n \geq n_0 $ for some $ n_0  \in \mathbb{N} $.

Also observe that $ \frac{b^{2n}}{b-1} = \underbrace{11 \dots 1}_{2n} . 111\dots $

$ \therefore \lim_{n \to \infty }\frac{ \frac{b^{2n}}{b-1}}{x_n} = 1 $

$ \Rightarrow \boxed{ \lim_{n \to \infty } \frac{b^n}{y_n} = \sqrt{b-1} }$

The crucial observation is $ \lim_{n \to \infty} ( by_n - y_{n+1} ) = \frac{b \sqrt{b-1}}{2} $, so let us proof it !

Claim : $ \lim_{n \to \infty} ( by_n - y_{n+1} ) = \frac{b \sqrt{b-1}}{2} $

$ b^2y_n^2 - y_{n+1}^2 = b^{n+2} + 3b^2 - 2b - 5 $

$ \Rightarrow (by_n - y_{n+1})( by_n + y_{n+1} ) = b^{n+2} + 3b^2 - 2b - 5 $

$ \Rightarrow (by_n - y_{n+1})\left( \frac{y_n}{b^n} + \frac{y_{n+1}}{b^{n+1}} \right) = b+ \frac{3}{b^{n-1}}-\frac{2}{b^n} - \frac{5}{b^{n+1}} $

$ \Rightarrow \boxed{ \lim_{ n \to \infty } ( by_n - y_{n+1} ) = \frac{b \sqrt{b-1}}{2} } $

Now $ by_n - y_{n+1} \in \mathbb{Z} \ \forall \ n \geq n_0 $

$ \therefore \frac{b \sqrt{b-1} }{2} \in \mathbb{Z} $, also as the sequence $ by_n - y_{n+1} \in \mathbb{Z} \ \forall \ n \geq n_0 $ therefore after a certain $M \in \mathbb{N} $ we have $ \boxed{ by_n - y_{n+1} = \frac{b \sqrt{b-1}}{2} \ \forall \ n \geq M } $.

$ \therefore 2by_n - b\sqrt{b-1} = 2y_{n+1} $ 

$ \Rightarrow b \ | \ 2y_{n+1} $

$ \Rightarrow b^2 \ | \ 4(b-1)y_{n+1}^2 $

$ \Rightarrow b^2 \ | \ 4( 3b - 5 ) $

Thus we have $ b^2 \leq 12b - 20 \Rightarrow (b-2)(b-10) \leq 0 $

Hence we deduce $5 \leq b \leq 10 $ simple case checking shows that $ b \ | \ 4(3b-5 ) $ only for $ b = 10 $.

Now if $ b = 10 $ we get $ x_n = \left( \frac{10^n + 5}{3} \right)^2 $ which clearly satisfies the problem condition.