National Math Olympiad – Final Round, 2015

For each $n \ge 2$ we can rearrange the given recursive formula to $\frac{a_{n+1}}{a_n} = -\frac{a_n}{a_{n-1}}$. Using this we calculate
$$\frac{a_{2n+2}}{a_{2n}} = \frac{a_{2n+2}}{a_{2n+1}} \cdot \frac{a_{2n+1}}{a_{2n}} = \left(-\frac{a_{2n+1}}{a_{2n}}\right) \cdot \left(-\frac{a_{2n}}{a_{2n-1}}\right) = \frac{a_{2n+1}}{a_{2n-1}}.$$
Using this equation twice in a row we get
$$\frac{a_{2n+2}}{a_{2n}} = \frac{a_{2n+1}}{a_{2n-1}} = \frac{a_{2n}}{a_{2n-2}},$$
which means that the sequence $a_2, a_4, a_6, \dots$ is geometric.

After we rearrange the formula to $\frac{a_{n+1}}{a_n} = -\frac{a_n}{a_{n-1}}$ we see that the sequence of positive real numbers $|a_1|, |a_2|, |a_3|, \dots$ is geometric, hence there exist positive real numbers $a$ and $q$ so that $|a_n| = aq^{n-1}$ for all positive integers $n$. It follows that for every $n$ we have either $a_n = aq^{n-1}$ or $a_n = -aq^{n-1}$. If we insert $2n+1$ instead of $n$ in the equation that the original sequence is defined with we get $a_{2n+2}a_{2n} = -a_{2n+1}^2 \le 0$, hence the neighbouring terms in the sequence $a_2, a_4, a_6, \dots$ are of opposite signs. We deduce that either
$$a_{2n} = (-1)^n a q^{2n-1} = -a q (-q^2)^{n-1}$$
for all $n$ or
$$a_{2n} = a q (-q^2)^{n-1}$$
for all $n$. In both cases the sequence $a_2, a_4, a_6, \dots$ is geometric.

We square the equality $a_n^2 = -a_{n+1}a_{n-1}$ to get $(a_n^2)^2 = a_{n+1}^2 a_{n-1}^2$. We rearrange this to $\frac{a_{n+1}^2}{a_n^2} = \frac{a_n^2}{a_{n-1}^2}$ and notice that the sequence of positive real numbers $a_1^2, a_2^2, a_3^2, \dots$ is geometric. Therefore there exist positive real numbers $a$ and $q$ such that $a_n^2 = aq^{n-1}$ for all positive integers $n$. It follows that for each $n$ we have either $a_n = \sqrt{a}\sqrt{q^{n-1}}$ or $a_n = -\sqrt{a}\sqrt{q^{n-1}}$. Similarly as in Solution 2 we get that either
$$a_{2n} = (-1)^n \sqrt{a} \sqrt{q^{2n-1}} = (-1)^n \sqrt{a} \sqrt{q} \sqrt{q^{2n-2}} = (-1)^n \sqrt{a q q^{n-1}} = -\sqrt{a q} (-q)^{n-1}$$
for all $n$ or
$$a_{2n} = \sqrt{a q} (-q)^{n-1}$$
for all $n$. In both cases the sequence $a_2, a_4, a_6, \dots$ is geometric.