National Math Olympiad

The numbers $a_1$ and $a_2$ are integers. Since $a_1=\frac{1}{\cos x}$, $a_2=\frac{1}{\cos (2x)}$ and $\cos (2x)=2(\cos x{)}^2-1$, we have $a_2=\frac{1}{2{\cos }^{2}x-1}=\frac{a_1^2}{2-a_1^2}$. Now, $a_2$ is an integer, so $2-a_1^2$ is a divisor of $a_1^2$. So, $2-a_1^2$ divides $2-a_1^2$ and $a_1^2$. It therefore also divides their sum, which is equal to 2. We conclude that $2-a_1^2$ is one of the numbers $-2,-1,1$ or $2$. In the first case we have $a_1^2=4$, in the second case we have $a_1^2=3$, in the third case we get $a_1^2=1$ and in the fourth case we get $a_1^2=0$. Since $a_1$ is a non-zero integer, we can only have $a_1=-2$, $a_1=2$, $a_1=-1$ or $a_1=1$. From here it follows that $x$ can only be equal to $0,\frac{\pi }{3},\frac{2\pi }{3},\pi ,\frac{4\pi }{3}$ or $\frac{5\pi }{3}$. If $x=0$, then $a_n=1$ for all $n$. If $x=\pi$, we get the sequence $-1,1,-1,1,\dots$. In the remaining four cases we have $a_{n+6}=a_n$, since for all integers $a$ we have $$\cos \left(\frac{(n+6)\cdot a\pi }{3}\right)=\cos \left(\frac{n\cdot a\pi }{3}+2a\pi \right)=\cos \left(\frac{n\cdot a\pi }{3}\right).$$ It therefore suffices to check that the first six terms of the sequence are integers. When $x=\frac{\pi }{3}$ and $\frac{5\pi }{3}$ we have $a_1=2$, $a_2=-2$, $a_3=-1$, $a_4=-2$, $a_5=2$, $a_6=1$. If $x=\frac{2\pi }{3}$ or $x=\frac{4\pi }{3}$, then the first six terms are $-2,-2,1,-2,-2,1$. The solutions are $0,\frac{\pi }{3},\frac{2\pi }{3},\pi ,\frac{4\pi }{3}$ and $\frac{5\pi }{3}$.