NZMO Round Two

Solution:
As long as $a_{n} \geqslant \sqrt{2}$ we have $a_{n + 1} = a_{n} - \sqrt{2}$ and so the sequence begins with an arithmetic progression with common difference $d = - \sqrt{2}$. Let $k$ be the first index such that $a_{k} < \sqrt{2}$. So $a_{k - 1} \geqslant \sqrt{2}$. Therefore
$$a_{k} = a_{k - 1} - \sqrt{2} \geqslant \sqrt{2} - \sqrt{2} = 0.$$
Thus we have $0 \leq a_{k} < \sqrt{2}$. From this we can calculate $a_{k} - \sqrt{2} < 0$ and so $\left|a_{k} - \sqrt{2}\right| = \sqrt{2} - a_{k}$.
$$\Longrightarrow a_{k + 1} = \left(\sqrt{2} -a_{k}\right)$$
Since $a_{k + 1} = \left(\sqrt{2} - a_{k}\right) \leqslant \left(\sqrt{2} - 0\right) = \sqrt{2}$, this means that $\left(a_{k + 1} - \sqrt{2}\right)$ is negative and thus
$$a_{k + 2} = \left|a_{k + 1} - \sqrt{2}\right| = \left(\sqrt{2} -a_{k + 1}\right) = \left(\sqrt{2} -\left(\sqrt{2} -a_{k}\right)\right) = a_{k}.$$
as required.