Winter Mathematical Competition

We prove that $\alpha$ is irrational. Suppose $\alpha$ is a rational number, i.e. the sequence $a_1,a_2,a_3,\dots ,a_k,\dots$ is periodic from some point onwards. Hence there exist $k_0$ and $T$ such that for any $k>k_0$ we have $a_k=a_{k+T}$.

Choose a positive integer $m$ for which $mT>k_0$ and $mT$ is a perfect square. If $T=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_s^{{\alpha }_s}$ then $m=p_1^{{\beta }_1}{p}_2^{{\beta }_2}\cdots p_s^{{\beta }_s}$, where ${\alpha }_i+{\beta }_i$ is even for all $i=1,2,\dots ,s$ and ${\beta }_i$ are large enough positive integers.

Choose a prime number $p>2007$, $p\ne p_i$, $i=1,2,\dots ,s$. Since $pmT-mT$ is divisible by $T$ we have that $a_{mT}=a_{pmT}$.

Denote by $\tau (k)$ the number of divisors of $k$ and by $f(k)$ the number of divisors of $k$ that are greater than $2007$. We have $f(pmT)=f(mT)+\tau (mT)$ and since $\tau (mT)$ is odd number we conclude that $f(pmT)$ and $f(mT)$ are of the same parity, a contradiction.