Hrvatska, 2011

Notice that $x_n > 1$, for all $n \in \mathbb{N}$. We also notice that all $x_n$ are rational so we can write $x_n = \frac{p_n}{q_n}$, where $p_n$ and $q_n$ are positive integers and $M(p_n, q_n) = 1$.

First let us prove that $p_n$ and $p_{n+1}$ are relatively prime for every $n \in \mathbb{N}$. We will prove that by induction. Obviously $M(p_1, p_2) = M(a, b) = 1$, i.e. $p_1$ and $p_2$ are relatively prime. Now we assume that $M(p_n, p_{n+1}) = 1$ for some $n$. Then
$$x_{n+2} = \frac{\frac{p_n^2}{q_n^2} + \frac{p_{n+1}^2}{q_{n+1}^2}}{\frac{p_n}{q_n} + \frac{p_{n+1}}{q_{n+1}}} = \frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{q_n q_{n+1} (p_n q_{n+1} + p_{n+1} q_n)} = \frac{p_{n+2}}{q_{n+2}}$$
Since $M(p_n, p_{n+1}) = 1$ by the inductive hypothesis and $M(p_{n+1}, q_{n+1}) = 1$, we conclude that $M(p_{n+1}, p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2) = M(p_{n+1}, p_n^2 q_{n+1}^2) = 1$, whence follows $M(p_{n+1}, p_{n+2}) = 1$. Thereby we have proved our assertion.

Now we want to prove that $x_n$ is not an integer for $n \ge 3$.
Assume the contrary, that $x_{n+2}$ is a positive integer for some $n \in \mathbb{N}$. Since
$$x_{n+2} = \frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{q_n q_{n+1} (p_n q_{n+1} + p_{n+1} q_n)} = \frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2}$$
we conclude that
$$q_{n+1} \mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \implies q_{n+1} \mid p_{n+1}^2 q_n^2 \implies q_{n+1} \mid q_n^2$$
because $p_{n+1}$ and $q_{n+1}$ are relatively prime. Now because of $q_{n+1} \mid q_n^2$ we have $q_{n+1}^2 \mid p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2 \implies q_{n+1}^2 \mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \implies q_{n+1}^2 \mid q_n^2$.

Analogously,
$$q_n \mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \implies q_n \mid p_n^2 q_{n+1}^2 \implies q_n \mid q_{n+1}^2$$
and then
$$q_n^2 \mid p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2 \implies q_n^2 \mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \implies q_n^2 \mid q_{n+1}^2$$

As $q_{n+1}^2 \mid q_n^2$ and $q_n^2 \mid q_{n+1}^2$, it follows that $q_n^2 = q_{n+1}^2$, that is $q_n = q_{n+1}$, and now we get
$$x_{n+2} = \frac{p_n^2 + p_{n+1}^2}{q_n (p_n + p_{n+1})}.$$
This means that
$$p_n + p_{n+1} \mid p_n^2 + p_{n+1}^2 \implies p_n + p_{n+1} \mid 2p_{n+1}^2$$
because
$$p_n^2 + p_{n+1}^2 = p_n^2 - p_{n+1}^2 + 2p_{n+1}^2 = (p_n - p_{n+1}) (p_n + p_{n+1}) + 2p_{n+1}^2.$$
Let $p$ be a prime number such that $p \mid p_n + p_{n+1}$, and thereby $p \mid 2p_{n+1}^2$.
If $p \neq 2$ then $p \mid p_{n+1}^2 \implies p \mid p_{n+1}$, and since $p \mid p_n + p_{n+1}$, it follows that $p \mid p_n$ which is a contradiction because $p_n$ and $p_{n+1}$ are relatively prime.
If $p=2$ is the only prime factor, then $p_n + p_{n+1}$ is a power of 2 bigger than 2 (because $p_n$ and $p_{n+1}$ are bigger than 1). It follows that $4 \mid p_n + p_{n+1}$ and then
$$4 \mid 2p_{n+1}^2 \implies 2 \mid p_{n+1}^2 \implies 2 \mid p_{n+1} \implies 2 \mid p_n.$$
which is again a contradiction since $M(p_n, p_{n+1}) = 1$.
Thereby we have proved that $x_n$ is not an integer for $n \ge 3$.