Hong Kong competition problems, 2016

$n$ can be any odd integer greater than $1$.

For even $n$, suppose $a_j + \frac{(-1)^j}{a_j} = b_j$ for $1 \le j \le n$, where $\{b_j\}$ is a permutation of $\{a_j\}$. The relation can be rewritten as
$$a_j^2 - a_j b_j = (-1)^{j+1}.$$
Summing over all $j$'s, we obtain
$$\sum_{j=1}^n (a_j^2 - a_j b_j) = 0$$
since $n$ is even. As $\{b_j\}$ is a permutation of $\{a_j\}$, this implies
$$\frac{1}{2} \sum_{j=1}^n (a_j - b_j)^2 = 0.$$
Hence, we must have $a_j = b_j$, which is impossible since $a_j + \frac{(-1)^j}{a_j} = b_j$. Therefore, there is no such sequence.

For $n=1$, we need $a_1 - \frac{1}{a_1} = a_1$, which is impossible.

So it remains to show that such a sequence exists when $n > 1$ is odd. First we construct a sequence $a_1, a_2, \dots, a_{n+1}$ such that $a_1 = 1 + x$ with $x > 0$, and
$$a_{j+1} = a_j + \frac{(-1)^j}{a_j}$$
for $1 \le j \le n$. It suffices to show $a_1, a_2, \dots, a_n$ are pairwise distinct, and choose $x$ such that $a_{n+1} = a_1$.

Note that
$$a_{j+2} = a_{j+1} + \frac{(-1)^{j+1}}{a_{j+1}} = a_j + \frac{(-1)^j}{a_j} + (-1)^{j+1} \left( a_j + \frac{(-1)^j}{a_j} \right)^{-1} = a_j + \frac{1}{a_j^3 + (-1)^j a_j}.$$
For odd $j$, since $a_1 > 1$, we can prove inductively that $a_{j+2} > a_j > 1$. Also, for even $j$, since $a_2 = a_1 - \frac{1}{a_1} > 0$, we can prove inductively that $a_{j+2} > a_j > 0$.

Therefore, if we can prove that $a_{n+1} = a_1$, then we have
$$0 < a_2 < a_4 < \dots < a_{n+1} = a_1 < a_3 < \dots < a_n,$$
and so all terms are pairwise distinct positive real numbers.

By using the recurrence relation, it is obvious that $a_{n+1}$ is a continuous function of $a_1$. When $a_1$ goes to $1$, we have
$$a_{n+1} - a_1 \ge a_4 - a_1 = a_2 + \frac{1}{a_2^3 + a_2} - a_1 \to \infty$$
since $a_2 \to 0$. This shows $a_{n+1} - a_1 > 0$ for some $x$.

On the other hand, for even $j$, as $a_j \ge a_2 = 1+x - \frac{1}{1+x} = \frac{2x+x^2}{1+x} > x$, we have
$$a_{j+2} = a_j + \frac{1}{a_j^3 + a_j} < a_j + \frac{1}{x^3},$$
and hence $a_{n+1} < a_2 + \frac{n-1}{2x^3}$ by summing over all $j$'s. It follows that
$$a_{n+1} - a_1 < a_2 + \frac{n-1}{2x^3} - a_1 = \frac{n-1}{2x^3} - \frac{1}{1+x} = \frac{(n-1) + (n-1)x - 2x^3}{2x^3(1+x)} < 0$$
for sufficiently large $x$.

Now, by the intermediate value theorem, there exists $x > 0$ such that $a_{n+1} = a_1$. So we are done.