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 $$q_{n+1}\mid p_n^2{q}_{n+1}^2+p_{n+1}^2{q}_n^2⟹q_{n+1}\mid p_{n+1}^2{q}_n^2⟹q_{n+1}\mid q_n^2$$ because $p_{n+1}$ and $q_{n+1}$ are relatively prime. Now because of $q_{n+1}\mid q_n^2$ we have $q_{n+1}^2\mid p_n{q}_n{q}_{n+1}^2+p_{n+1}{q}_{n+1}{q}_n^2⟹q_{n+1}^2\mid p_n^2{q}_{n+1}^2+p_{n+1}^2{q}_n^2⟹q_{n+1}^2\mid q_n^2$.

Analogously, $$q_n\mid p_n^2{q}_{n+1}^2+p_{n+1}^2{q}_n^2⟹q_n\mid p_n^2{q}_{n+1}^2⟹q_n\mid q_{n+1}^2$$ and then $$q_n^2\mid p_n{q}_n{q}_{n+1}^2+p_{n+1}{q}_{n+1}{q}_n^2⟹q_n^2\mid p_n^2{q}_{n+1}^2+p_{n+1}^2{q}_n^2⟹q_n^2\mid q_{n+1}^2$$

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 $$p_n+p_{n+1}\mid p_n^2+p_{n+1}^2⟹p_n+p_{n+1}\mid 2p_{n+1}^2$$ 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 $p\ne 2$ then $p\mid p_{n+1}^2⟹p\mid p_{n+1}$, 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 $$4\mid 2p_{n+1}^2⟹2\mid p_{n+1}^2⟹2\mid p_{n+1}⟹2\mid p_n.$$ 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 $n\ge 3$.